Solitons

September 29, 2006

• General references: Drazin and Johnson - this course organzed a bitdifferently

• http://www.maths.dur.ac.uk/˜dma0saa/Solitons

• First term: general ideas about solitons+ some specific methods (Back-lund transforms and Hirota method

• Secnd term: A powerful general technique called inverse scattering. De-veloped in late 1960’s

1 Introduction

1.1 What is a soliton?

‘Experimental’ discovery by John Scott-Russell in 1834 on the Forth ClydeCanal. “A rounded smooth well-defined heap of water”, going along at about8mph. Different heights → different velocities.

Definition: A soliton is a solution to a p.d.e. which is

1. localized,

2. keeps its localized form over time

3. is preserved under interactions with other solitons

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Nowadays can use the relevant equations for water waves and solve numericallyto show these properties, as above. These were plotted using the program inthe appendix A in mathematica. The Korteweg de Vries equation is the onewe need (extension from the simplest linearized equation which is valid forinfinitessimally small displacements)

ut + 6uux + uxxx = 0 (1)

It is instructive to see what solitons are ‘made of’ by looking at whathappens if we remove certain terms from the KdV equation. The first optionis to eliminate the non-linear term;

ut + uxxx = 0 (2)

The result is dispersion (waves spread and energy disperses);

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The second option is to remove the dispersive term and leave onlynon-linearity

ut + 6uux = 0

The result is breaking (waves energy concentrates until it becomes singular);

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A soliton is a stable balance of these two concentrating and dispersingeffects. Not all non-linear p.d.e.s admit such solutions. Those that do arecalled “integrable”. Property 3 usually implies that there is 1 space and onetime dimension.

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1.2 Relevance of solitons

• Applied maths - water waves, optical fibres etc.

• particle physics - models of elementary particles like protons, neutronsetc.

• “pure” mathematics; particularly since the most recent discoveries, stud-ies of solitons involves many different and developing areas

1.3 Relevance to inverse scattering problems

Strictly speaking the word Soliton (coined by Kruskal and Zabrusky in 1965)refers to situtations where the soliton forms part of a more complicated evo-lution involving maybe two or more solitons plus possibly plain wave com-ponents. (A single isolated wave is called a solitary wave although I won’tbother making a distinction.) If we begin with an initial configuration that isnot a solitary wave then soon the different components begin to emerge. Forexample consider the evolution of the following initial gaussian form of u on aperiodic interval;

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The initially the wave radiates plane waves and the soliton content emerges.Looking at the final configuraiton (where the soliton has gone once roundthe periodic interval) it is clear that the soliton is a peristent component.There could have been two or more solitons emerging depending on the initalconfiguration. The decomposition of functions into these components will betreated in the 2nd term.

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2 Waves, Dispersion and Dissipation

The usual situation for a localized water wave is that it speads out as it travels- it disperses. This prevents it from being a soliton. Need to understand thisphenomenon first. Consider some examples;

1. Wave equation (linear)1

v2utt − uxx = 0 (3)

D’Alembert’s solution is u(x, t) = f(x − vt) + g(x + vt). Superpositionof left and right moving waves - trivially shape preserving

2. First order linear equation

1

vut + ux = 0 (4)

Now u = f(x − vt) and again all waves move with fixed speed v. Nodispersion.

3. Klein-Gordon equation

1

v2utt − uxx +m2u = 0 (5)

Try the solution u = ei(kx−ωt). Get a dispersion relation, ω =v√k2 +m2. k is the wave-number and ω is the frequency. The ve-

locity of the wave crests is c = ω/k so that different wave-numbers moveat different speeds.

If we make a lump by superposition of different wavelengths,

u(x, t) =

∫ ∞

−∞f(k)ei(kx−ωt)dk, (6)

then it will disperse.

• Phase-velocity; c(k) = ω(k)/k; so for the above example have c(k) =

v√k2+m2

k.

• Group-velocity; cg(k) = dωdk

. This is roughly the speed the lump u(x, t)moves at while it disperses. Proof a bit messy - consider an example ina minute.

NB : for KG equation, have c > v but cg = v k√k2+m2 < v. In QM the KG

equation can describe a particle moving at speed cg. The speed of lightis v, so even though c > v relativity is OK because the signal is movingat cg.

NB2 : lump with varying k↔uncertainty principle. In words, smaller parti-cles (δx) need larger average k, and so δk is bigger.

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2.1 An example of group velocity; the Gaussian wavepacket

f(k) = e−a2(k−k)2 (7)

Let u = Re(z) where

z(x, t) =

∫ ∞

−∞e−a

2(k−k)2ei(kx−ωt) dk

Most of u comes from k ≈ k, so expand ω(k) near k.

ω(k) ≈ ω(k) +dω

dk(k − k) + . . .

= ω(k) + cg(k − k) + . . .

so that

z(x, t) ≈∫ ∞

−∞e−a

2(k−k)2ei(kx−t(ω(k)+cg(k−k))) dk

= ei(kx−ω(k)t)

∫ ∞

−∞e−a

2(k−k)2ei(k−k)(x−cgt) dk

= ei(kx−ω(k)t)

∫ ∞

−∞e−a

2k2

ei(k(x−cgt) dk

= ei(kx−ω(k)t) e−1

4a2 (x−cgt)2∫ ∞

−∞e−a

2(k− i2a2 (x−cgt))2 dk

=

√π

aei(kx−ω(k)t) e−

14a2 (x−cgt)2

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The first factor is a ‘travelling’ plane wave part. The second piece is an enve-lope centred on x− cgt, with width 2a. To see the dispersion need to expandto higher order in k − k. An example numerical solution is shown above forthe KG equation at time t = 0 and later time (the dispersion, which must golike d2ω/dk2, is clearly visible).

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To see the dissipation expand ω(k) to higher order ;

ω(k) ≈ ω(k) +dω

dk(k − k) + ω′′(k − k)2 + . . .

= ω(k) + cg(k − k) + χ(k − k)2

where χ = ω′′(k). Now

z(x, t) =

∫ ∞

−∞e−(a2+iχt)(k−k)2ei(kx−t(ω(k)+cg(k−k))) dk (8)

so we can simply replace a with

b(t) =√a2 + iχt (9)

and repeat previous steps to arrive at

z(x, t) =

√π

bei(kx−ω(k)t) e−

14b2

(x−cgt)2 (10)

To see the dissipation we can use an envelope function |z| noting that

|z| ≥ Re(z) = u (11)

it is not difficult to show

|z| =√

π

|b|ei(kx−ω(k)t) e

− 14|b|2

(x−cgt)2 (12)

where

|b| = a

(1 +

χ2t2

a4

) 12

(13)

So now the envelope function is a gaussian hump of width 2|b| that spreadsout over time.

2.2 Examples with dissipation

If ω is not real we get dissipation.

• First order dissipation;

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vut + ux + αu = 0 (14)

where α is real. Here ω(k) = v(k − iα) so not real. The plane wavesolutions are

u = eik(x−vt)e−vαt (15)

If α > 0 the amplitude decays to zero as t → ∞(dissipation). Nodispersion here.

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• The heat equationut − αuxx = 0 (16)

Here have ω(k) = −iαk2 and a solution is

u = eikxe−αk2t (17)

Again dissipation.

2.3 Summary

• Various linear equations describing waves in 1-dimension

• dispersion relation ω(k).

• If ω complex have can have dispersion or dissipation.

• Both generally break up waves.

• A different behaviour if k is imaginary is evanescence . The wave dieswith distance (common in QM tunneling phenomena).

• Solitons are not of this type and they don’t disperse - non-linearity iscrucial ingredient.

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3 Travelling waves

In this section we’ll find some travelling wave solutions such as the KdV soliton.Travelling waves are solutions of the form u(x, t) = f(x− vt). i.e. some fixedprofile travelling at constant speed v. If the solution is localized it’s part wayto being a soliton. The thing we have to check is the preserved form underinteraction (property 3). So solutions of utt − uxx = 0 are trivial in the sensethat there is no interaction.

3.1 KdV soliton

ut + 6uux + uxxx = 0 (18)

Substituting u = f(x− vt) gives

−vf ′ + 6ff ′ + f ′′′ = 0 (19)

which we can integrate once to give

−vf + 3f 2 + f ′′ = A (20)

where A is a constant. Multiply by f ′ and integrate again gives

−1

2vf 2 + f 3 +

1

2(f ′)2 = Af +B (21)

with B another constant. Now apply boundary conditions. The wave issolitary, so at x→ ±∞ we need it and its derivatives to vanish;

f |±∞ = f ′|±∞ = f ′′|±∞ = 0

givingA = B = 0 (22)

Then (f ′)2 = f 2(v − 2f) or∫df

f√v − 2f

= ±(x− x0 − vt) (23)

Note I have reabsorbed the integration constant into the definition of x, sothat wave begins at x = x0 when t = 0.

Try a substitution f = 12v sech 2θ so that df = −v sech 2θ tanh θ dθ (since

d(cosh−1) = − cosh−1 tanh) so that

df

f√v − 2f

= − v sech 2θ tanh θdθ12v sech 2θ

√v tanh θ

= − 2√vdθ (24)

where I used (1−sech 2 = tanh2). Hence − 2√v

∫dθ = −2θ/

√v = ±(x−x0−vt)

and

u(x, t) = f(x− vt) =v

2sech 2 (

√v

2(x− x0 − vt) (25)

The ± is irrelevant but the velocity is positive, v ≥ 0. The waves have

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speed vheight v

2

width 1√v

Three examples are shown below with speeds v = 1, 2, 3.

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NB. Solutions for non-zero A,B involve Jacobi theta functions. (see D+Jchapter 2). Actually the figures on the first page are only good uptox = ±40 rather than x = ±∞ (computers can’t handle the latter verywell) so they are really Jacobi theta functions but close to the real soliton.(I used periodic boundary conditions at |x| = 40, so these really areperiodic functions. This is more obvious in the dispersive diagrams wherethe small ripples reappear to the right of the bounding interval.)

NB2. In the limit of small u amplitude the KdV equation linearizes andyou expect standard trig functions. The above periodic theta functionsapproximate to the expected trig functions in this limit.

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3.2 Solitons in the Sine-Gordon equation

uxx − utt = sin u (26)

Again substituting u = f(x− vt) and using uxx = f ′′ and utt = v2f ′′ we get

(1− v2)f ′′ = sin f (27)

or more succinctlyf ′′ = γ2 sin f (28)

where γ = 1/√

1− v2. (This looks relativistic which is not a coincidence. TheSG eqn has a relativistic form . The speed of light = 1.) Now multiply by f ′

and integrating (using f ′ sin f = −(cos f)′) gives

(f ′)2 = A− 2γ2 cos f (29)

orf ′ = ±

√A− 2γ2 cos f. (30)

At x = ±∞ we expect f ′′ = 0 since we are looking for an isolated lump offinite extent, so

f(±∞) = 2πn (31)

where n ∈ Z. This also tells us from f ′(±∞) = 0 that

A = 2γ2. (32)

Integrating eq.30 we have∫df√

1− cos f= ±

√2γ(x− vt)

√2 log tan(f/4) = ±

√2γ(x− vt) (33)

So thatu(x, t) = 4 tan−1

(e±γ(x−x0−vt)

). (34)

This is called a kink or antikink. Three kinks are shown below with speedsv = 0.5, 0.9, 0.99 (getting steeper respectively). (Note the step is through 2π).

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3.3 Aside: Connection with particle kinematics

There is a fantastically useful way to picture soliton solutions. Go back toeq.21 with A = B = 0 and rearrange a bit;

1

2(f ′)2 + U(f) = 0 (35)

whereU(f) = −v

2f 2 + f 3 (36)

This equation is identical to that of a particle of mass 1 rolling in the potentialU(f) with ξ playing the roll of time (the first term is the Kinetic energy thesecond term the potential). The potential is shown below;

A B

Consider the ξ → −∞ limit. Here we require f ′ = f = 0 so that the totalenergy is zero. The particle begins stationary at f = 0 at point A. After sittingfor a semi-infinite period of time it rolls down the slope and by conservationof energy it reaches point B, and then rolls back again to A and comes to restat ξ → +∞. B corresponds to the top of the soliton since f ′ = 0, and wesee immediately that fmax = v/2 since U(fmax) = 0 simply by conservationof energy. This method is extremely useful for visualizing solitons for moregeneral functions.

For the Sine Gordon equation the potential is

U(f) = γ2(cos f − 1). (37)

The difference in this case is that the particle rolls from one peak to anotheras ξ goes from −∞ to +∞.

NB: The U(f) here is the negative of the potential energy V in the equationsof motion; this is because we are letting x play the roll of time and theusual equation of motion is of the form utt − uxx − V ′(u) = 0

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3.4 Physical model for the SG soliton

Consider the physical set-up ...

a

l

θnm

consisting of masses m separated by elastic of length a on pendula of length l.If θn is the angle of the nth pendulum then by resolving forces and accelerationin radial direction (exercise) we get

ml2θn = −mgl sin θn +k

a(θn+1 − θn) +

k

a(θn−1 − θn) (38)

where k is a coefficient of force in the elastic (linear by Young - related toYoung’s modulus of elasticity). Now let our x direction be given by na = x sothat if a is small can approximate

k

a(θn−1 + θn+1 − 2θn) = ka

(θn+1 − θn)/a− (θn − θn−1)/a

a≈ ka θ′′(x) (39)

So eqn of motion becomes

ml2θ = −mgl sin θ + ka θ′′ (40)

Now let a→ 0 and k →∞ such that ka→ ml2 is finite. Then

θtt − θxx = −gl

sin θ (41)

which is the SG equation.The system has zero energy (is in the “vacuum”) when θ = 0∀x. Two

other possibilities; small oscillations as shown in the figure above and “kinks”as below where θ(−∞) = 0 and θ(+∞) = 2π

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m

In terms of θ this is

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Θ

The one twist configuration can’t be deformed back into the “vacuum” in acontinuous way (unlike the small wave configuration which can). This is atopological property (see next lecture).

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3.5 A topological defect that is not a soliton: φ4 theory

Consider a one-dimensional field theory (like the SG theory) but with theequation of motion

utt − uxx = λu(a2 − u2) (42)

This (c.f. appendix B.4) is equivalent to a field theory in a potential

P(u) =λ

4(u2 − a2)2

known as the φ4 potential (conventionally φ is used for scalar field theories).Again substituting u = f(x− vt) and using uxx = f ′′ and utt = v2f ′′ we get

f ′′ = γ2λf(f 2 − a2) (43)

and then integrating up gives eq.458 in the appendix. Taking the square rootof eq.458 we get

f ′ = ±γ√λ

2(f 2 − a2). (44)

Again at ξ = ±∞ we expect both f ′ and f ′′ = 0 so that the integrationconstant has been set to zero and

f(±∞) = ±a. (45)

The potential looks like

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so the particle kinematic equivalent of the would-be soliton is a particle sta-tionary at −a at ξ = −∞ rolling to +a as ξ → +∞ in the inverted potential.Integrating we have∫

df

(a2 − f 2)= ±

√λ/2γξ

u = ±a tanh(√

λ/2 aγ(x− vt))

(46)

A stationary example (with λ = 2 and a = 1) is shown below. Again thetheory is relativistic so the solution depends on γ(x− vt).

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Why is this not a soliton? Because it doesn’t satisfy invariance underinteraction. Below I show before and after shots of two interactions that I didnumerically; the first is between two SG solitons and the second interactionbetween two of the φ4 would-be solitons. The latter looks almost like a solitoncollision, but you can just make out small oscillations. The φ4 collision isslightly “inelastic” with some kinetic energy of the walls being transferred toradiation (oscillations of the scalar field).

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4 Topological lumps and the Bogomolny bound

Take SG as exampleutt − uxx + sinu = 0 (47)

(Note that we can always remove g/l by rescaling t and x).

KE = T =

∫ ∞

−∞

1

2(ut)

2 dx

PE = V =

∫ ∞

−∞

1

2(ux)

2 + (1− cosu) dx (48)

Note that in terms of Lagrange density

L =1

2u2t −

1

2u2x − (1− cosu) = T − V (49)

so that the Euler Lagrange eqn

d

dt

∂L∂ut

+d

dx

∂L∂ux

− ∂L∂u

= 0 (50)

indeed give the SG equation eq.47. The form of V (i.e. the “1”) comes fromthe requirement of finite energy so that V → 0 at x → ∞ where u → 2πn.(The pendula must all be pointing down at infinity).

• Physically the absolute number n is irrelevant - a shift u(x) → u(x) +2πk doesn’t change any of the equations in particular V(u). What isimportant is the difference which measures the number of kinks

u(+∞)− u(−∞) = 2π(m− n) (51)

This number is invariant under a finite 2πk shift. The number n−m istopological because unchanged under continuous changes to u. e.g.

u(x) → u(x) + f(x)

changes n−m only if f has non-zero n−m.

• If n−m 6= 0 then u can never dissipate to zero.

• Localized non-dissipative solutions that arise this way are topologicallumps. So the SG kinks are topological lumps and also solitons - (i.e.still have to prove property 3).

• The conservation of n−m is a topogical conservation law

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4.1 The Bogomolny bound

The kink solution we found earlier was static. But are such solutions (local)minima of the energy and therefore stable? Bogomolny’s argument show’s theyare. In many field theories (e.g. electromagnetism) the Bogomolny boundmakes finding topologically non-trivial solutions much easier. For example“complete the square” of the energy for the SG kink

E = T + V =

∫ ∞

−∞

1

2u2t +

1

2u2x + (1− cosu) dx

≥∫ ∞

−∞

1

2u2x + (1− cosu) dx

=

∫ ∞

−∞

1

2u2x + 2 sin2

(u2

)dx

=

∫ ∞

−∞

1

2

(ux ± 2 sin

(u2

))2

∓ 2 sin(u

2

)ux dx

=

∫ ∞

−∞

1

2

(ux ± 2 sin

(u2

))2

dx ± 4[cos

(u2

)]∞−∞

(52)

If u is a single kink then

4[cos

(u2

)]∞−∞

= −8

and we get the Bogomolny bound E ≥ 8 with saturation when

ux − 2 sin(u

2

)= 0 (53)

Solving this ∫dx =

∫du

2 sin(u2

) = log tan(u

4

)(54)

so thatu = 4 tan−1 ex−x0

Note that this is the earlier solution in the rest frame of the soliton

x− x0 → γ(x− x0 − vt)

we asked for the static solution so this makes sense. (Also note there is clearlya problem if there is more than 1 kink - the Bogomolny argument breaksdown.)

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In fact we can extend this to generic relativistic scalar field theories usingAppendix B4.

E =

∫T + V dx =

∫ ∞

−∞

1

2u2t +

1

2u2x + P(u) dx

≥∫ ∞

−∞

1

2u2x + P dx

=

∫ ∞

−∞

1

2(ux ±

√2P)2 ∓

√2Pux dx

=

∫ ∞

−∞

1

2(ux ±

√2P)2 dx ∓

[∫ √2Pdu

]u(∞)

u(−∞)

(55)

When saturated, this inequality tells us that

u2x

2− P(u) = 0 (56)

which is simply eq.458 when the wave is static (γ = 1) and A = 0. It also tellsus the total energy of this configuration is the Bogomolny bound energy;

Ekink = ±[∫ √

2Pdu]u(∞)

u(−∞)

(57)

4.2 The energy is localized...

For the static SG kink, the energy density is

1

2u2x + 2 sin2

(u2

)= u2

x

= 2 sech2 (x− x0) (58)

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4.3 Derrick’s Theorem:

So far we have concentrated on solutions in d = 1 space dimensions. Is it possi-ble to get static stable lumps (solitonic or not) in 2 or more space dimensions?The answer (for any number of simple scalar fields such as displacement) isno;

Derrick’s Theorem: Let φ be a set of n scalar fields assembled into ann-vector, and let the dynamics be defined by

L = φt.φt − φx.φx − φy.φy . . . − P(φ)

Then for d ≥ 2 the only time independent non-singular solutions of finiteenergy are the ground states.

Proof: Define

K =1

2

∫ddx (∇φ)2

P =

∫ddxP(φ)

(Note I am avoiding using T and V since V includes the (∂xφ)2 termsin the rest of these notes.) Assume that both K and P must be non-negative (we do not allow the possibility of negative P.E.) and that P =K = 0 only for the groundstate. Now, if there exists a solution φ(x),then there also exists a one parameter family of configurations φ(λx) ofwhich only λ = 1 is the true stable solution. The energy of this familyof configurations can be related to the energy of the true solution by asimple rescaling of the integration variable x

E = λ(2−d)K + λ−dP

Clearly all values of λ have the same topological properties so we canuse Bogomolny’s argument; the true solution minimizes the total energyand therefore

d

dλ(K + P )λ=1 = 0 = (2− d)K − dP (59)

If d > 2 we must have P = K = 0. If d = 2 the above equationimplies only the vanishing of P and a small amount of further argumentis needed; if P vanishes ∀x, it is constant and hence

∇P = ∇φ.∂P∂φ

= 0 ∀x

Either P (φ) = constant (which would be trivial) or it follows that ∇φ =0∀x and hence K = 0.

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4.4 Summary

To be acceptable lumps must be stable. This can happen in two ways, inte-grability or topology;

KdV SG

true solitons topological lumps

φ4

The φ4 model lumps are topological because the model has a Z2 symmetry in

P(φ) =λ

4(φ2 − a2)2.

The Z2 symmetry is invariance of the action under φ → −φ.(See first exer-cise sheet.) However they are not true solitons because they radiate duringcollisions.

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5 Backlund Transformations

So far we have considered only isolated solitary wave solutions. How can wedescribe two interactions (after all this defines one of the essential propertiesof solitons)? We need another trick called the Backlund transformation .They have the following uses

• Generating solutions to difficult p.d.e.s from known solutions to easierones.

• Generating new solutions to p.d.e.s from already known solutions.

The second of these (called an auto-Backlund transformation) is the onewe will need to generate two and more soliton solutions from the solitary wavesolutions we know.

5.1 Definition

Suppose that we write down an arbitrary pair of p.d.e. relations which I’lllabel with a star;

R1[u, v] = 0

R2[u, v] = 0. F

These relations can be non-linear and any order, and act on two arbitraryfunctions which I’ll call u and v. Now if we can eliminate u and v separatelyfrom these relations we will find two separate p.d.e.s which I’ll label with aheart and spade;

P [u] = 0 ♥Q[v] = 0 ♠

(P [u] is an entire differential equation now, not to be confused with the po-tential energy). Let ♥ be an equation such as the SG equation, and let u bea solution of it. Then the relations F generate a solution v to ♠ from each u.

Why is this useful? Because the equations F and ♥ may be easier to solvethan ♠.

• the relations F are collectively called a Backlund transformation.

• if P = Q then it’s an auto-Backlund transformation and we get newsolutions to the same p.d.e.

21

5.2 Example 1: Auto-BT for the Laplace equation

Take

R1[u, v] ≡ ux − vy = 0

R2[u, v] ≡ uy + vx = 0 (60)

Differentiating R1 by x and R2 by y and adding we get P [u], and differentiatingR2 by x and R1 by y and subtracting we find Q[v];

P [u] ≡ uxx + uyy = 0

Q[v] ≡ vxx + vyy = 0 (61)

Example: u(x, y) = 2xy is a simple solution of P [u] = 0. Then F (eq.60) gives

vx + 2x = 0

vy − 2y = 0 (62)

Integrating givesv = y2 − x2, (63)

a different solution to the Lapace equation. In this case R1 and R2 are theCauchy-Riemann conditions for a function v + iu to be analytic. In this casethe function is v+ iu = (x+ iy)2. P [u] = Q[v] = 0 is the statement that boththe real and imaginary parts satisfy Laplace’s equation.

NB: To go beyond just flipping between two solutions we have to introducean extra parameter into auto-BT’s.

NB2: Given u there are two equations in relation F for only one function v.Normally this would be overdetermined and not have any solutions. e.g.if u = x2 then F gives vx = 0, and vy = 2x which is inconsistent. Butu = x2 is not a solution to P [u] = 0, so P is an...

Integrability Condition: The relations F only consistently generate a func-tion v from a given u provided that P [u] = 0.

I presented BT’s this way round (beginning with F and deriving both P andQ) so that the condition is more evident. Often as in D+J one begins with Pand looks for F to get Q.

22

5.3 Example 2: BT’s for the Liouville equation

The Liouville equation isv+− = ev. (64)

It is similar to the SG equation which could be written as

v+− =1

2i(eiv − e−iv)

however unlike SG it has no static solutions because it has no minima of thepotential. Here I am using light-cone coordinates described in Appendix Cand in particular have used eq.464 to gett he above.

Eq.64 is the equation we want to arrive at as our Q[v]. Consider thefollowing BT

∂+(v − u) =√

2e(v+u)/2

∂−(v + u) =√

2e(v−u)/2 (65)

First cross-differentiate

∂−∂+(v − u) =√

2∂−e(v+u)/2

=1√2e(v+u)/2∂−(v + u)

= ev (66)

where I used the second BT relation in the last line. Doing the same for thesecond relation yields

∂+∂−(v + u) = ev (67)

Adding and subtracting eqs.66 and 67 gives

u+− = 0

v+− = ev (68)

These are our P [u] and Q[v] respectively. The former is the linear wave equa-tion, the latter is Liouville’s equation.

Conclusion: Any solution for the linear wave equation can gen-erate via eq.65 a solution to the Liouville equation. Converselyall solutions to Liouville’s equation are related to a solution to thelinear wave equation.

23

5.4 General solution for the Liouville equation

The above BT generates solutions to the Liouville equation from the waveequation. However for the wave equation we know a general solution due toD’Alembert, so we must be able to generate general solutions for Liouville aswell. According to D’Alembert

u = f(x+) + g(x−) (69)

is a general solution for the linear wave equation (light-cone coordinates makethis obvious).

Each F equation becomes an o.d.e. for v that we have to solve to get x+

and x− dependence of v; consider

∂+(v − u) =√

2e(v+u)/2 (70)

Trick: substituting v − u = −2 logψ this eqn. becomes

2

ψ∂+ψ = −

√2e− logψ+u =

√2

ψeu.

∂+ψ = − eu√2

(71)

and integrating we get

ψ(x+, x−) = − 1√2

∫ x+

eu(z+,x−)dz+ + C(x−) (72)

where C(x−) is a constant of z+ integration. Now we can insert our (or ratherD’Alembert’s) solution of eq.69 for u;

ψ = e(u−v)/2

= e(f(x+)+g(x−))/2e−v/2 (73)

so that

e−v/2 = −e−(f(x+)+g(x−))/2

√2

∫ x+

ef(z+)+g(x−)dz+ + C(x−)e−(f(x+)+g(x−))/2

= −e−(f(x+)−g(x−))/2

√2

∫ x+

ef(z+)dz+ + C(x−)e−(f(x+)+g(x−))/2

= −e−(f(x+)−g(x−))/2

√2

F (x+) + C(x−)e−(f(x+)+g(x−))/2 (74)

where I defined

F (x+) =

∫ x+

ef(z+)dz+. (75)

Now repeat the exercise with the second BT relation: we put u+v = −2 logψand get

∂−ψ = − 1√2e−u

24

and hence

e−v/2 = −e−(f(x+)−g(x−))/2

√2

G(x−) +D(x+)e(f(x+)+g(x−))/2 (76)

where

G(x−) =

∫ x−

e−g(z−)dz− (77)

and D(x+) is another constant of integration. All that remains is to determineC and D. Comparing eqs.74 and 76 we find

C(x−) = −e+g(x−)

√2

G(x−)

D(x+) = −e−f(x+)

√2

F (x+) (78)

and then eq.74 gives

e−v/2 = − 1√2e(g(x−)−f(x+))/2(F (x+) +G(x−)) (79)

or

v(x+, x−) = f(x+)− g(x−)− 2 log

(−F (x+) +G(x−)√

2

)(80)

This is the general solution to Liouville’s equation. (first found by Liouvillein 1853). Most of the time we can’t get a general solution like this. It’s fairlytrivial to verify this is a solution; indeed

∂+∂−v =2F ′G′

(F +G)2

ev = ef−g2

(F +G)2.

Example: Of course this solution is general, but to demonstrate, take a verysimply linear solution ax+ + bx− of the linear wave equation with a andb constant. We have

f(x+) = ax+

g(x−) = bx−

F (x+) =eax+

a+ c

G(x−) =e−bx−

b+ d

(c, d constant) and hence a considerably less simple

v = ax+ − bx− − 2 log

(−be

ax+ + aebx−√2ab

+ const

)

25

One final point is that there is a one parameter family of BT’s given by

∂+(v − u) = a√

2e(v+u)/2

∂−(v + u) =1

a

√2e(v−u)/2 (81)

Any value of a is equally good as it yields the same P and Q equations. Thiswill be important in cases where we cannot get the full solution later.

26

6 The Sine-Gordon equation

u+− = sinu (82)

Try the BT

∂+(u− v) =2

asin

(1

2(u+ v)

)∂−(u+ v) = 2a sin

(1

2(u− v)

)(83)

where a is a constant parameter. Again cross-differentiate

∂+∂−(u− v) =1

acos

(1

2(u+ v)

)∂−(u+ v)

= 2 cos

(1

2(u+ v)

)sin

(1

2(u− v)

)= sinu− sin v (84)

and

∂+∂−(u+ v) = a cos

(1

2(u− v)

)∂−(u− v)

= 2 cos

(1

2(u− v)

)sin

(1

2(u+ v)

)= sinu+ sin v. (85)

Adding these two and subtracting them gives respectively

u+− = sinu

v+− = sin v. (86)

so this is an auto-BT for the SG equation.

6.1 Generating SG kinks from the vacuum

Let us begin with the solution u = 0 which is trivally a solution to SG (it’sthe vacuum). The BT gives us

u+ =2

asin

u

2

u− = 2a sinu

2(87)

These can be easily integrated using∫du

sin u2

= 2 log tanu

4(88)

27

giving

2

ax+ = 2 log tan

u

4+ g(x−)

2ax− = 2 log tanu

4+ f(x+) (89)

with f and g being two “constant” of integration functions. Subtracting wefind

2

ax+ + f(x+) = 2ax− + g(x−). (90)

Since either side depends on different variables each side must be constant(like sep’n of variables in p.d.e.s) which I’ll take to be −2c, so that

f(x+) = −2

ax+ − 2c

g(x−) = −2ax− − 2c. (91)

Hence

2 log tanu

4=

2

ax+ + 2ax− + 2c (92)

oru = 4 tan−1

(e

1ax++ax−+c

). (93)

Finally we need to convert back to regular x, t coordinates;

1

ax+ + ax− = x(

1

2a+a

2) + t(

1

2a− a

2)

=1 + a2

2a

(x− a2 − 1

a2 + 1t

). (94)

If we define

v =a2 − 1

a2 + 1

then it is easy to verify that

1 + a2

2|a|=

1√1− v2

= γ

as required and thatu = 4 tan−1

(e±γ(x−vt)+c

), (95)

with the ± (kink or antikink) being chosen by the sign of a and direction ofmotion being left or right for |a| < 1 and |a| > 1 respectively; we have

a < −1 −1 < a < 0 0 < a < 1 1 < a

Right moving anti-kink Left moving anti-kink Left moving kink Right moving kink

The BT “creates” a kink or anti-kink from the vacumm solution (i.e. v = 0).The nature of it depends on parameter a.

28

Remarkable fact: this is true more generally! The BT has the effect ofadding kinks or antikinks to any solution (even if it has dispersive waves).i.e. if v has kink number n and we solve F for u then u(x, t) will havekink number n± 1 depending on a.

Direct proof?: is left as a challenge (I don’t know of one)! Indirect proofsbased on inverse scattering method (see next term) have been aroundsince 1975.

So we can think of the BT as a “machine”/”black box” to add kink/antikinksolutions.

We can use the BT repeatedly to add kinks and anti-kinks whereever we like(depending on the constants of integration c) and with whatever velocity welike; for example

adds three kinks to the solutions u0 which may be the vacuum.

Problem: The integrations get progressively harder (although still easier thanSG). Try for example getting the 2 kink solution from the one kink one!Fortunately we can use a theorem.

29

6.2 The theorem of Permutability (Bianchi 1902) andmulti-kink solutions

General observation: Multiple solitons have no static solution. To see thisgo back to the particle rolling in the inverted potential picture; If it takesan infinite “time” ξ for the particle to roll into the well, then it takes aninfinite time to roll out again by symmetry.

Imagine doing two successive BT’s with parameters a1 and a2;

in two different orders as shown. u3 and u4 both look like u0 with two solitonsadded.

Theorem: For any u1 and u2 it is possible to choose constants of integrationin the 2nd BT’s such that u3 = u4.

In terms of the diagrams we have the above. The a1 and a2 BTs have beenarranged to commute (this depends on choosing the constants of integrationcorrectly).

Now consider just the ∂+ part of the BTs and look at the upper route

30

We get

∂+(u1 − u0) =2

a1

sin1

2(u1 + u0)

∂+(u3 − u1) =2

a2

sin1

2(u1 + u3). (96)

Adding gives

∂+(u3 − u0) =2

a1

sin1

2(u1 + u0) +

2

a2

sin1

2(u3 + u1) (97)

Taking instead the lower route

we get

∂+(u2 − u0) =2

a2

sin1

2(u2 + u0)

∂+(u3 − u2) =2

a1

sin1

2(u2 + u3). (98)

and adding gives

∂+(u3 − u0) =2

a2

sin1

2(u2 + u0) +

2

a1

sin1

2(u3 + u2) (99)

Equating the two expressions for ∂+(u3−u0) gives us a simple algebraic relationbetween the solutions;

2

a1

sin1

2(u1 + u0) +

2

a2

sin1

2(u3 + u1) =

2

a2

sin1

2(u2 + u0) +

2

a1

sin1

2(u3 + u2).

(100)Given the vacuum and the one-kink solutions we can always solve this to getthe two-kink solutions. We can bootstrap our way to large kink number with-out doing any integrating at all! Sometimes called a non-linear superpositionprinciple (tells us how to put solutions together).

As a check we should get the same relation if we use the ∂−(u+ v) part ofthe BT instead. From the top route we get

∂−(u1 + u0) = 2a1 sin1

2(u1 − u0)

∂−(u3 + u1) = 2a2 sin1

2(u3 − u1). (101)

31

Subtracting gives

∂−(u3 − u0) = 2a2 sin1

2(u3 − u1)− 2a1 sin

1

2(u1 − u0). (102)

From the bottom route we get

∂−(u2 + u0) = 2a2 sin1

2(u2 − u0)

∂−(u3 + u2) = 2a1 sin1

2(u3 − u2). (103)

and again subtracting gives

∂−(u3 − u0) = 2a1 sin1

2(u3 − u2)− 2a2 sin

1

2(u2 − u0) (104)

Equating the two expressions for ∂−(u3 − u0) gives

2a2 sin1

2(u3−u1)−2a1 sin

1

2(u1−u0) = 2a1 sin

1

2(u3−u2)−2a2 sin

1

2(u2−u0).

(105)What we now need to do is check that eqs.(100) and (105) are the same. UsingsinA± sinB = 2 sin 1

2(A±B) cos 1

2(A∓B) we can rewrite eq.100 as follows;

1

a1

(sin1

2(u1 + u0)− sin

1

2(u3 + u2)) =

1

a2

(sin1

2(u2 + u0)− sin

1

2(u3 + u1))

2

a1

(sin1

4(u1 + u0 − u2 − u3) =

2

a2

(sin1

4(u2 + u0 − u3 − u1)

× cos1

4(u1 + u0 + u3 + u2)) × cos

1

4(u2 + u0 + u3 + u1))

2a2(sin1

4(u1 + u0 − u2 − u3) = 2a1(sin

1

4(u2 + u0 − u3 − u1) (106)

where I cancelled the common cosine factor. Likewise we can rewrite eq.105as

a2(sin1

2(u3 − u1) + sin

1

2(u2 − u0)) = a1(sin

1

2(u3 − u2) + sin

1

2(u1 − u0))

2a2(sin1

4(u3 + u2 − u1 − u0) = 2a1(sin

1

4(u1 + u3 − u2 − u0)

× cos1

4(u3 + u0 − u2 − u1)) × cos

1

4(u3 + u0 − u2 − u1))

2a2(sin1

4(u3 + u2 − u1 − u0) = 2a1(sin

1

4(u1 + u3 − u2 − u0) (107)

i.e. the same equation so it is consistent.These equations are still a bit inconvenient because we want to get u3

from u0 (the vacuum say) via the intermediate stage single kinks u1 and u2.Manipulate eq.107; let A = u0−u3 and B = u1−u2. Then this equation reads

a2 sin1

4(A+B) = a1 sin

1

4(A−B)

32

or

a2(sinA

4cos

B

4+ sin

B

4cos

A

4) = a1(sin

A

4cos

B

4− sin

B

4cos

A

4).

Dividing through by cos B4

cos A4

we get

(a1 − a2) tanA

4= (a1 + a2) tan

B

4

or

tan1

4(u0 − u3) =

(a1 + a2)

(a1 − a2)tan

1

4(u1 − u2) (108)

6.3 Two soliton solutions to SG

Begin with u0 = 0 (the vacuum). Using the identity tan(A+B) = tanA+tanB1−tanA tanB

we get

tanu3

4=

(a2 + a1)

(a2 − a1)

tan u1

4− tan u2

2

1 + tan u1

4tan u2

2

.

Recall that the expression for the single kinks has

tanui=1,2

4= eθi

where

θi =x+

ai+ aix− + ci

ε(ai) γ(ai) (x− x0 − v(ai) t) (109)

where the integration constant c1 has determined the position of the kink x0

and ε(a) = Sign(a). Then we get

u3 = 4 tan−1

((a2 + a1)

(a2 − a1)

eθ1 − eθ2

1 + eθ1+θ2

). (110)

Note that as expected when we try to get a static solution we have v = 0 forboth kinks or a1 = ±a2 and the solution breaks down.Two soliton solutions toSG

6.4 3 soliton solution

Just to show how we can now “bootstrap” to any number of kinks let’s look atthe 3 kink solution. Begin with u0 as a single kink solution parameterized bya0 and u1,2 as two of the two-kink solutions parameterized by adding kinks a1,2

respectively. Then to get to u3 we must add the complementary a2,1. Usingthe previous relation we have

tan(u3 − u0)

4=

(a2 + a1)

(a2 − a1)

tan u1

4− tan u2

2

1 + tan u1

4tan u2

2

.

33

Substituting in the tan(u/4) expressions for the two-kink solutions we get(using ti = tan(ui/4))

t3 − t01 + t0t3

=a2 + a1

a2 − a1

a0+a1

a0−a1

(t1−t01+t0t1

)− a0+a2

a0−a2

(t2−t01+t0t2

)1 + a0+a1

a0−a1

(t1−t01+t0t1

)a0+a2

a0−a2

(t2−t01+t0t2

)

which can be easily solved for t3. It looks a bit lob-sided because one of thesingle kinks is on the LHS, but with a bit of effort the solution is seen not tobe. To write it I’ll define

a±±± = (a0 ± a1)(a1 ± a2)(a2 ± a0)

where the ± refer to the respective positions inside the brackets. The solutionis

tanu3

4=

a+−+eθ0 + a++−e

θ1 + a−++eθ2

a+−+eθ1+θ2 + a++−eθ0+θ2 + a−++eθ0+θ1 − a−−−(111)

6.5 Asymptotics of interactions

First note that our observation of no static solutions was correct. A staticsolution has a1 = a2 = ±1 so that u3 = 4 tan−1(∞) or 4 tan−1(0) both ofwhich are constant and sitting in a vacuum. The solution break down. In factthere are no solutions where two kinks have the same speed a1 = ±a2 (this isbecause you could always Lorentz boost to the rest frame which would haveto be a static solution). So there must always be an interaction.

We now want to look at the asymptotics of interactions to make sure soli-tons are preserved. The two kink solutions should contain “hidden” one kinksolutions as t → ±∞. An “after” shot of a kink/antikink solution is shownbelow;

-20 -15 -10 -5 5 10 15

2

4

6

8

10

The kink/antikink pair look preserved. To see this in the solution take t →t1 + δt with δx = x − x0 − v1t1 where v1 =

a21−1

a21+1

and let t1 → ±∞ so we are

looking around the location of the first kink.

34

We have

θ1 = ε1γ1(δx− v1δt)

θ2 = ε2γ2(δx− v1δt) + ε2γ2(v1 − v2)(t1 + δt) (112)

Near the a1 kink we have (by definition) δx, δt ≈ 0, so that θ1 is small butθ2 → ±∞.

t→ −∞ t→ +∞v2 < v1 θ2 → −ε2∞ θ2 → +ε2∞v2 > v1 θ2 → +ε2∞ θ2 → −ε2∞

Defining sign(θ2) = −η2 the asymptotic form of the solution around δx = 0can be written

limt→±∞

(u3) = 4 tan−1

(a2 + a1

a2 − a1

η2eη2θ1

). (113)

But defining a “phase shift”

c1 =1

γ1

loga2 + a1

a2 − a1

(114)

we can rewrite this as

limt→±∞

(u3) = 4 tan−1(η2e

η2(θ1+η2γ1c1)). (115)

Finally we can use −1/ tan x = tan(x− π/2) with x = tan−1 eθ1+η2c1 to get

limt→±∞

(u3) =

{4 tan−1

(e(θ1+γ1c1)

)η2 > 0

−2π + 4 tan−1(e(θ1−γ1c1)

)η2 < 0

(116)

We can recognize three features

1. The form of the soliton is preserved - it remains a kink or antikink

2. There is a “phase shift” in the kink location of x0 → x0 + 2c1 as η2 goesfrom negative to positive.

3. There is always a drop or rise of 2π in this kink between asymptotict→ ±∞ (due to the interaction with the other kink)

35

-10-5

0

5

10

0

510

1520

-5

0

5

-10-5

0

5

0

510

1520

-10 -5 0 5 10

5

7.5

10

12.5

15

17.5

20

-10

-5

05

10

05

1015

20

-5

0

5

-5

05

10

510

1520

-10 -5 0 5 10

5

7.5

10

12.5

15

17.5

20

The above figure shows the asymptotics in two ways. The second way is acontour plot of u2

x which remember is a measure of the energy stored in thefield.

A subtle point:

Examine eq.113 a bit more closely for a particular example. Say we wantto have a kink labelled “1” coming from large negative x, travelling to therightward and interacting with a kink labelled “2” travelling left. Going toour table of kinks and antikinks we would take

a1 > 1 1 > a2 > 0

But now we get a bit of a surprise; at t→ −∞ kink-1 is actually an antikinkbecause a2+a1

a2−a1< 0 and ε2 = +1 and so η2 = +1 for these values! (The same

reversal happens if we begin with an antikink with a1 < −1).Not only that but look at the 2nd soliton. We now have to use the θ2 table

the other way around using the v2 > v1 row but with 1 ↔ 2. So here we havean η1 which is negative as t→ −∞. The two negatives cancel and this solitonremains what you’d expect. The conclusion is that BT’s don’t necessarilyrespect the “kinkness” or “antikinkness” of a solution, but just add ±1 kinkwithout caring much what kind it is.

36

Interpretation as force

There is an interaction that looks like kink/kink repulsion and kink/antikinkattraction. The force gets infinite as the speeds become equal and zero -another way to think of no static solutions.

This similarity with electric charge explains the importance of this type ofinteracting field theory for particle physics. Promoted in the 60’s by Skymeand useful for e.g. describing nuclear interactions. This suggests that theremay be bound states of kink/antikink pairs as well by comparison with whathappens in nuclear models. It turns out there are, called breathers.

7 The breather solution

Look for a bound solution in the 2 soliton solution. Recall

u3 = 4 tan−1

((a2 + a1)

(a2 − a1)

eθ1 − eθ2

1 + eθ1+θ2

). (117)

For any a1 and a2 the permutability theorem guarantees it’s a solution of SG,even if ai are complex. But u3 should be real (the field theory is defined withreal fields representing e.g. displacement angles). Options leading to real u3;

1. a1, a2 (and also c1 , c2) ∈ R. already seen this

2. a2 = a∗1(and also c2 = c∗1) now try this

To show option 2 gives real u3recall θi = x+

ai+ aix− + ci, so that

θ2 = θ∗1

and

u∗3 = 4 tan−1

((a∗2 + a∗1)

(a∗2 − a∗1)

eθ∗1 − eθ

∗2

1 + eθ∗1+θ∗2

)= 4 tan−1

((a1 + a2)

(a1 − a2)

eθ2 − eθ1

1 + eθ1+θ2

)= u3. (118)

Let’s look for a solution involving arbitrary a1 = a. Define for shorthand

a = reiϑ

θ1 = α+ iβ

so that

tanu3

4=

2 cosϑ

−2i sinϑ

eα+iβ − eα−iβ

1 + e2α

= −cosϑ

sinϑ

sin β

coshα. (119)

37

So now need to determine α, β. We have

α = Re(θ1)

= (1

rx+ + rx−) cosϑ+Re(c)

=1 + r2

2rcosϑ (x− x0 −

r2 − 1

r2 + 1t)

= cosϑ γ(r) (x− x0 − v t) (120)

where γ(r) and v(r) are the same definitions as earlier and I absorbed Re(c)into x0. We also have

β = Im(θ1)

= (−1

rx+ + rx−) sinϑ+ Im(c)

=1 + r2

2rsinϑ (

r2 − 1

r2 + 1(x− x0)− t)

= sinϑ γ(r) (v(x− x′0)− t) (121)

So then

tanu3

4=

cosϑ

sinϑ

sin(sinϑ γ(t− vx))

cosh(cosϑ γ(x− vt))(122)

where I took x0 = x′0 = 0 for simplicity. So the denominator is an envelopefunction that moves along at speed v = r2−1

r2+1. Inside the envelope there is an

oscillation with faster phase velocity 1/v. (Note the speed of light =1 so thatv < 1).

To check out the wobble choose r = 1so that v = 0and then

tanu3

4=

cosϑ

sinϑ

sin(sinϑ t)

cosh(cosϑx)(123)

We can always “boost” this solution to a v 6= 0 frame. The fields values andenergy density look like a bouncing kink/antikink pair as below. The figure isfor ϑ = π/20. As ϑ→ 0 the kink/antikink are more loosely bound, get furtherapart during the bounce and the bounce takes longer. The bounce period andsize are

τ =2π

sinϑλ =

1

cosϑ

respectively when static, and the solutions all look like the figure if the x andt axes are rescaled appropriately. Notice that at speed the bounce size is

2π

cosϑγ

which shortens due to Lorentz contraction.

38

-5

-2.5

0

2.5

5 0

20

40

60

-2

-1

0

1

2

-5

-2.5

0

2.5

5

-6 -4 -2 0 2 4 6

0

10

20

30

40

50

60

70

Much more “phenomenology” can be done with these solutions which are amodel for many processes in particle physics.

39

8 BT’s for the KdV equation

We now consider another example of BT’s for the KdV equation. In thefollowing section we will again use the theorem of permutability to get multi-soliton solitons, so we are following exactly the same procedure as for the SGequation. BT’s here are a bit (OK, a lot) more involved. The SG BT wasknown in the 1880’s, whereas the KdV BT had to wait until 1973 (Wahlquistand Estabrook).

Supposeu = λ− v2 − vx (124)

satisfies the KdV equation, where λ is constant (it will eventually be the freeparameter we need in the BT) and v = v(x, t). After a bit of effort we findthat this implies that

(2v +∂

∂x)(vt + 6(λ− v2)vx + vxxx) = 0 (125)

so that if v satisfies(vt + 6(λ− v2)vx + vxxx) = 0 (126)

it implies that u given by the above is a solution of the KdV equation.

Proof: KdV equation for u is ut + 6uux + uxxx = 0. So that

0 = −2vvt − vxt + 6(λ− v2 − vx)(−2vvx − vxx)− 2vvxxx − 6vxvxx − vxxxx

= −(2v +∂

∂x)vt − 6(λ− v2)(2v +

∂

∂x)vx + 12vv2

x + 6vxvxx − 2vvxxx − 6vxvxx − vxxxx

= −(2v +∂

∂x)(vt + 6(λ− v2)vx)− 6

∂v2

∂xvx + 12vv2

x − 2vvxxx − vxxxx

= −(2v +∂

∂x)(vt + 6(λ− v2)vx + vxxx) (127)

Eq.173 with the value λ = 0 is called the modified KdV equation (mKdV) -c.f. question 1 on sheet 2, and u = −v2 − vx is called a Miura transformation(found by R.Miura in 1968). Now if v satisfies eq.173 then clearly so does−v.So if

u1 = λ− v2 − vx

u2 = λ− v2 + vx (128)

then u1,2 both satisfy the KdV equation. This is the first step. The idea isnow to construct an autoBT linking u1 and u2 going via the mKdV equation.

From the above we can rearrange to get

u1 − u2 = −2vx

u1 + u2 = 2(λ− v2) (129)

40

It’s convenient to define w1,2 such that

ui =∂wi∂x

so then the two equations become

w1 − w2 = −2v

(w1 + w2)x = 2λ− 1

2(w1 − w2)

2 (130)

where I already substituted for v in the second. This is the “x” part of theBT. The “t” part comes from the mKdV equation itself substituting v. Wehave

0 = −1

2(w1 − w2)t + 6(λ− v2)(−1

2(w1 − w2)x)−

1

2(w1 − w2)xxx

= (w1 − w2)t + 3(w1 + w2)x((w1 − w2)x) + (w1 − w2)xxx

= (w1 − w2)t + 3(w21,x − w2

2,x) + w1,xxx − w2,xxx (131)

So collecting the two important equations for our BT

(w1 + w2)x +1

2(w1 − w2)

2 − 2λ = 0 (132)

(w1 − w2)t + 3(w21,x − w2

2,x) + w1,xxx − w2,xxx = 0 (133)

So to repeat the point of BT’s, if we have one w1 differentiating to a particularKdV solution, u1, then this BT ensures that w2 differentiates to a u2 which isalso a KdV solution, (because the relevant v satisfied the mKdV equation).

8.1 Check;

The obvious check is to start with the Baclund transformation and to makesure it derives the KdV equaiton for both u1 and u2. In terms of wi the KdVequation is

wi,xt + 6wi,xwi,xx + wi,xxxx = 0 i = 1, 2 (134)

Differentiating eq.133 by x gives immediately

(w1 − w2)xt + 6(w1,xw1,xx − w2,xw2,xx) + w1,xxxx − w2,xxxx = 0 (135)

so if we can find a similar equation but with + signs everywhere we can subtractand add and get our KdV equations. The obvious step is to differentiate 132with respect to t . To simplify things I’ll define A = w1 +w2 and B = w1−w2.The BT becomes

Ax +B2

2− 2λ = 0

Bt + 3AxBx +Bxxx = 0 (136)

41

Differentiating the first of these by t gives

0 = Axt +BBt

= Axt −B(3AxBx +Bxxx)

where I substituted from the 2nd half of the BT in the last line. Now weneed to get Axxxx. We can keep differentiating the first equation to get someidentities;

Axx +BxB = 0 (137)

Axxx +BxxB +B2x = 0 (138)

Axxxx +BxxxB + 3BxBxx = 0 (139)

Adding Axxxx from the last of these gives

Axt + Axxxx − (3AxBBx +BBxxx + Axxxx) = Axt + Axxxx − (3AxBBx − 3BxBxx)

= Axt + Axxxx + 3(AxAxx +BxBxx)

Finally we are ready to expand out and find (sinceAxAxx+BxBxx = 2(w1,xw1,xx+w2,xw2,xx)) that

(w1 + w2)xt + 6(w1,xw1,xx + w2,xw2,xx) + w1,xxxx + w2,xxxx = 0 (140)

In conjunction with eq.135 (adding and subtracting) we see both u1 = w1,x

and u2 = w2,x satisfy the KdV eqn.

8.2 The KdV solitary wave from the vacuum

Choose w2 = 0 and for convenience of notation write w1 ≡ w. Then eq.132implies

wx = 2λ− 1

2w2∫

dw

4λ− w2=

x

2+

1

2A(t) (141)

where A(t) is a constant of x integration. Now find

1

2√λ

tanh−1(w/2√λ) =

x

2+A

2

w = 2√λ tanh(

√λ(x+ A(t)) (142)

if λ > 0 (otherwise would find tan). Now need to fix A(t) using eq.133.Reading it off we have

wt + 3w2x + wxxx = 0 (143)

42

Differentiating eq.132 once we get relations wxx = −wwx and wxxx = −w2x −

wwxx = −w2x + w2wx, which we can use to simplify. We have

0 = wt + 2w2x + w2wx

= wt + 2wx(wx +1

2w2)

= wt + 4λwx (144)

So this equation gives (∂

∂t+ 4λ

∂

∂x

)w = 0 (145)

which means simply that w is a function of x−x0−4λt where x0 is a constant,and we can write

w(x, t) = 2√λ tanh(

√λ(x− x0 − 4λt)) (146)

To get the previous solution we identify v = 4λ (this is velocity v not to beconfused with the v in the mKdV equation) and we get

w =√v tanh(

√v

2(x− x0 − vt))

Finally we need to differentiate to get u which gives

u = wx =v

2sech 2(

√v

2(x− x0 − vt)),

the solitary wave solution for KdV.Note that there is another solution that we will need later; if |w| > 2

√λ

then the integral would have given

1

2√λ

coth−1(w/2√λ) =

x

2+A

2

w = 2√λ coth(

√λ(x− x0 − 4λt)) (147)

This solution is singular when the argument vanishes (i.e. where 1/tanh = 0).

8.3 The theorem of Permutability and non-linear su-perposition for KdV

Again we use that fact that for any pair of single soliton solutions w1 and w2

it is possible to choose constants of integration in the 2nd BT’s such that asubsequent BT yields the same w4.

43

Now we can consider just the ∂x part of the BTs (remember the two parts gavea consistency check in the case of SG, but here the ∂t part of the BT wouldbe a bit complicated to do). The upper route gives

(w0 + w1)x = 2λ1 −1

2(w0 − w1)

2

(w1 + w3)x = 2λ2 −1

2(w1 − w3)

2 (148)

The lower route gives

(w0 + w2)x = 2λ2 −1

2(w0 − w2)

2

(w2 + w3)x = 2λ1 −1

2(w2 − w3)

2 (149)

Subtracting the second from the first equation in both cases gives

(w3 − w0)x = 2(λ2 − λ1) +1

2(w0 − w1)

2 − 1

2(w1 − w3)

2

= 2(λ1 − λ2) +1

2(w0 − w2)

2 − 1

2(w2 − w3)

2 (150)

Equating these gives

4(λ2 − λ1) =1

2(w0 − w2)

2 +1

2(w1 − w3)

2 − 1

2(w0 − w1)

2 − 1

2(w2 − w3)

2

= (w3 − w0)(w2 − w1)

so that

w3 = w0 +4(λ2 − λ1)

w2 − w1

(151)

As before have a purely algebraic relation between the solutions. This isanother non-linear superposition principle.

44

8.4 Two soliton solutions

As for the SG we can now build a two soliton solution for the KdV equation.One interesting feature is that immediately from eq.151 we can see that toavoid singularities we need to avoid w1 = w2. But two non-singular tanhsolutions for wi always cross whereas we can choose one singular and onenon-singular solution so they do not cross, as shown below.

-2 2 4 6

-2

-1

1

2

-2 2 4 6

-6

-4

-2

2

4

6

The singular solution has to be the one with the larger velocity. The result isa non-singular w3.

To get the solution define

θi =√λi(x− x0 − 4λit) (152)

Then the two solutions are assuming λ2 > λ1

w1 = 2√λ1 tanh θ1

w2 = 2√λ2 coth θ2

and since w0 = 0 is the vacuum, we have

w3 =2(λ2 − λ1)√

λ2 coth θ2 −√λ1 tanh θ1

=2(λ2 − λ1) sinh θ2 cosh θ1√

λ2 cosh θ1 cosh θ2 −√λ1 sinh θ1 sinh θ2

Finally we need to differentiate this to get u3. The resulting function isn’tparticularly pleasant,

u3 = 2(λ2 − λ1)λ2 cosh2 θ1 + λ1 sinh2 θ2(√

λ2 cosh θ1 cosh θ2 −√λ1 sinh θ1 sinh θ2

)2 (153)

For the asymptotics we can do the same operation as for the SG equation. Lookaround the solution labelled i = 1 or 2. Take t→ ti+δt with δx = x−x0−vitiwhere vi = 4λi and let ti → ±∞. We have

θi =

√vi2

(δx− viδt)

θj 6=i =

√vj

2(δx− vjδt) +

√vj

2(vi − vj)ti (154)

45

Near the i soliton we have (by definition) δx, δt ≈ 0, so that θi is small butθj 6=i → ±∞.

t→ −∞ t→ +∞i = 1 θj=2 → −∞ θj=2 → +∞i = 2 θj=1 → +∞ θj=1 → −∞

Defining sign(θj) = ηj the asymptotic form of the solution around δx = 0 canbe written by noting that cosh θj → 1

2e|θj | and sinh θj → η

2e|θj |. We get

u3|i = 2(λ2 − λ1)λi(√

λ2 cosh θi −√λ1ηj sinh θi

)2 (155)

We now use√λ2 cosh θi −

√λ1ηj sinh θi =

1

2(√λ2 − ηj

√λ1)e

θi +1

2(√λ2 + ηj

√λ1)e

−θi

=

√(√λ2 − ηj

√λ1)(

√λ2 + ηj

√λ1)

2×

√(√λ2 − ηj

√λ1)√

(√λ2 + ηj

√λ1)

eθi +

√(√λ2 + ηj

√λ1)√

(√λ2 − ηj

√λ1)

e−θi

(156)

=

√(λ2 − λ1)

2

√

(√λ2 − ηj

√λ1)√

(√λ2 + ηj

√λ1)

eθi +

√(√λ2 + ηj

√λ1)√

(√λ2 − ηj

√λ1)

e−θi

(157)

Now define a phase shift

c =1

2log

√(√λ2 −

√λ1)√

(√λ2 +

√λ1)

(158)

We have√λ2 cosh θi −

√λ1ηj sinh θi =

√(λ2 − λ1)

2

(eθi+ηjc + e−(θi+ηjc)

)=

√λ2 − λ1 cosh(θi + ηjc) (159)

and substituting we get

u3|i =vi2

sech 2(

√vi2

(x− x0 − vit+2ηj√vic).

Going to our table, we see that the slower i = 1 wave is shifted back (i.e. x0

gets a negative contribution) ) by

4√vic

46

and the faster i = 1 wave shifted forwards by a slightly smaller amount butotherwise they retain their form. Note we could have done the asymptoticsusing w3 rather than u3 which might have been a bit easier. The result in a3D plot in the (x, t) plane looks like

-5

0

5

10

15

-2

0

2

4

0

2

4

6

-5

0

5

10

and three stages in the interaction look like

-20 -10 10 20

1

2

3

4

5

6

7

-20 -10 10 20

1

2

3

4

5

6

7

-20 -10 10 20

1

2

3

4

5

6

7

47

9 Current/charge conservation

Current conservation is an important part of many aspects of physics. Forsolitons they help to explain why the motion is so restricted that it can becomputed exactly. The basic idea is to construct quantities, some combinationof u, ux, uxx..., ut, utt... that are time constant during for any solution;

Q =

∫F (u, ux, uxx..., ut, utt...) dx

such thatdQ

dt= 0.

An example for the KdV equation is

Q1 =

∫u dx

which tells us that the total area under the solution is constant. Clearly thisrestricts the solution. We have already met the topological kink number inthe SG equation, but it will turn out that the KdV equation has an infinitenumber of conserved charges restricting the flow. This means that not much(e.g. chaos) can happen. The fact that solitons do not break up under collisioncan be put down to the conservation of charges.

How do we derive this and other conserved charges? Consider the conser-vation of mass for gas in a pipe flowing at speed V (x) along its length. In anelement δx long gas flow in at a rate ρ(x)V (x) where ρ is the density, and out

at a rate ρ(x + δx)V (x + δx) = ρ(x)V (x) + δx∂(ρV )∂x

. The nett mass outflow

from the element is δx∂(ρV )∂x

. But the rate of loss of mass is −∂(δxρ)∂t

. If mass isconserved then the two must be equal and

∂ρ

∂t+∂(ρV )

∂x= 0 (160)

This is the mass conservation equation and restricts possible velocity flow.What happens to the total mass M =

∫ρ(x)dx over time?

d

dt

∫ ∞

−∞ρ dx =

∫ ∞

−∞

∂ρ

∂tdx

= −∫ ∞

−∞

∂(ρV )

∂xdx

= −[ρV ]∞−∞= 0 (161)

if the density or velocity is zero (or even just equal) at the endpoints at x→±∞. Total mass in the pipe is conserved.

48

9.1 Generalization of this idea

Assume two quantities X and T can be assembled from the u ux uxx . . . suchthat

∂T

∂t+∂X

∂x= 0 (162)

For the example above the role of X was played by ρu the density flow, and Twas the local density. Assume further that X remains constant at x→ ±∞

X → c (163)

The eq.162 means that

d

dt

∫ ∞

−∞T dx =

∫ ∞

−∞

∂T

∂tdx

= −∫ ∞

−∞

∂X

∂xdx

= −[X]∞−∞= 0 (164)

so

Q =

∫ ∞

−∞T dx

is a conserved quantity (it was the total mass in the previous example).

9.2 Example: the wave equation

Consideru+− = 0 (165)

Note that any polynomial h of u− obeys ∂+h(u−) = 0 since

∂+h(u−) = h′u+− = 0.

Furthermore since ∂+ = ∂t + ∂x we can set

X = T = (u−)n (166)

for any n ∈ Z+. Similarly we can also have

X = −T = (u+)n (167)

Setting n = 1 gives nothing since using D’Alembert’s solution u = f(x+) +g(x−) we get

Q =

∫ ∞

−∞g′ dx

=

∫ ∞

−∞gx dx = 0

49

Setting n = 2 we get two independent quantities

(ux ± ut)2 = u2

x ± 2uxut + u2t

Equivalently we can assemble them into more recognizable combinations

E =

∫ ∞

−∞

1

2(u2

t + u2x) dx

M =

∫ ∞

−∞utux dx (168)

E is the total energy conservation in the absence of a potential P(u). M is themomentum. Clearly we get an infinite number of conserved quantities herebut then it’s somewhat trivial.

9.3 Conserved currents in the KdV equation

Looking at the KdV equation

ut + 6uux + uxxx = 0

it is obvious that the equation itself is in the form eq.162 with

T = u

X = 3u2 + uxx (169)

so that

Q1 =

∫ ∞

−∞u dx

must be conserved. That is the area under any solitary waves or interactingsolitons is preserved as long as the interval is infinite. Now multiply KdV byu to get

1

2(u2)t + (2u3 + uuxx −

1

2u2x)x = 0 (170)

so that remarkably

Q2 =

∫ ∞

−∞u2 dx

is constant as well. As you can imagine the possibilities for u are gettingincreasingly restricted. The next one is more tricky. We have to do (3u2 −ux

∂∂x

)×KdV to get

(u3 − 1

2u2x)t + (some function)x = 0 (171)

Q3 =

∫ ∞

−∞u3 − 1

2u2x dx

50

Physically (in e.g. water waves) these three conservation laws correspond toconservation of mass, momentum and energy. But the series does not stopthere. By trial and error conserved currents were found upto

Q10 =

∫ ∞

−∞(u10 − 60u7u2

x − {29 terms}+1

4862(uxxxxxxxx)

2 dx

(found in the 1960’s) where Qn =∫∞−∞(un + . . .) dx. This gets increasingly

messy, but a general method was discovered that shows there are an infinitenumber of conserved currents.

9.4 The Gardner transform (1968)

This argument begins with the generalized mKdV seen earlier by transformingu as follows

u = λ− v2 − vx (172)

to get

(2v +∂

∂x)[vt + 6(λ− v2)vx + vxxx] = 0 (173)

At the moment notice that although solutions v to the mKdV (the stuff in[ ]’s) are also solutions to KdV, the converse is not necessarily true. Gardnermodified this equation by putting

v = εw +1

2ε

λ =1

(2ε)2(174)

Then

λ− v2 = −w − ε2w2

u = −w − εwx − ε2w2 (175)

The extra parameter ε will help us generate the currents. We now find

(2εw +1

ε+

∂

∂x)[εwt − 6(w + εw2)εwx + εwxxx] = 0. (176)

Hence u given byu = −w − εwx − ε2w2

is a solution of the KdV equation if w satisfies

[wt − 6(w + εw2)wx + wxxx] = 0. (177)

We see immediately that setting ε = 0 we get the KdV equation (albeit withreversed sign in the middle term). Again this equation is in the form Tt+Xx =0 and hence

Q =

∫ ∞

−∞w dx

51

is conserved for any ε.Any solution w at non-zero ε will be a function of ε and so let’s expand

the w asw =

∑n

wnεn. (178)

w0 is some solution that is valid when ε = 0 so it satisfies the KdV equation(with sign reversed on the wwx term . Our conserved charge can also beexpanded

Q =

∫ ∞

−∞w dx

=∑n

∫ ∞

−∞wnε

n dx

=∑n

Qnεn (179)

Since ε is a free parameter the Qn must each individually be conserved so wehave an infinite number of conserved currents. (For example, Q0 is obviouslyconserved since we can always take ε = 0. But now the combined Q−Q0 mustalso be conserved. Again taking ε→ 0 we see that limε→0(Q− Q0)/ε = Q1 isconserved, and so on.)

All we need to find the Qn are the functions wn. From the above we have

u = −w − wx − ε2w2

= −∑n

wnεn −

∑n

wn,xεn+1 − ε2(

∑n

wnεn)2 (180)

We now simply have to iteratively equate coefficients of ε to find the wn interms of u. In detail

w0 = −uw1 + w0,x = 0

w2 + w1,x + w20 = 0

w3 + w2,x + 2w0w1 = 0

w4 + w3,x + w21 + 2w0w2 = 0 (181)

and so on. Then substituting u and working down the ladder we finally get

w0 = −uw1 = ux

w2 = −(uxx + u2)

w3 = uxxx + 2uux

w4 = −w3,x − u2x − 2u(u2 + uxx) etc. (182)

52

You then get wn=odd are total derivatives so∫wn=odd dx doesn’t tell us much.∫

w0dx = −Q1∫w1dx = 0∫w2dx = −Q2∫w3dx = 0∫w4dx = −2Q3 etc (183)

Note we can essentially ignore w3,x part in the w4 integral because it is a totalderivative. (To get the last one you have to use uuxx = (uux)x − u2

x).

53

10 Symmetries and conservation laws

There is an important and very deep connection between symmetries andconservation laws. Symmetries can be seen most easily in the lagrangian for-malism;

S[u] =

∫ t2

t1

∫ ∞

∞L(u, ut, ux) dxdt (184)

with for example

L =u2t

2− u2

x

2− (1− cosu) (185)

Hamilton’s principle is that if we let u → u + δu then if δS = 0 implies theEuler-Lagrange equations. One thing we neglected before though, which isimportant now are the boundary terms. Making the variation we actually get

δS = 0 =

∫ ∫L(u+ δu, ut + δut, ux + δux)− L(u, ut, ux) dtdx

=

∫ ∫∂L∂u

δu+∂L∂ut

δut +∂L∂ux

δux dtdx

=

∫ ∫ (∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

)δu dtdx

+boundary terms (186)

(Here for example d/dt means use the chain rule with u(x, t) and ux(x, t) butdo not differentiate with respect to x.) Setting everything in brackets to zerogives the E-L equation as earlier.

∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

= 0 (187)

However to get the last two terms I integrated by parts once and generatedsome “boundary terms” (i.e. complete derivatives). In other words I used

d

dx

(∂L∂ux

δu

)= δu

d

dx

(∂L∂ux

)+ δux

(∂L∂ux

)d

dt

(∂L∂ut

δu

)= δu

d

dt

(∂L∂ut

)+ δut

(∂L∂ut

)(188)

and the additional boundary terms are∫ ∫d

dx

(∂L∂ux

δu

)+d

dt

(∂L∂ut

δu

)dxdt.

These will be important for telling us about current conservation later butsince they are total derivatives they make no difference to the equations ofmotion. For the above example we get the SG equation back from the E-Lequation;

utt − uxx + sinu = 0.

54

10.1 Additional Lagrangians 1: the KdV equation

Now consider

L =1

2wxwt + w3

x −1

2w2xx. (189)

This is now a function of wxx as well as w, wx and wt so we need to modifythe Euler-Lagrange equation slightly. At the risk of confusion I’ll still useu to discuss the Euler-Lagrange equation. So again let u → u + δu then ifS → S + δS we extend fairly easily what was done before

δS = 0 =

∫ ∫L(u+ δu, ut + δut, ux + δux, uxx + δuxx)− L(u, ut, ux, uxx) dtdx

=

∫ ∫∂L∂u

δu+∂L∂ut

δut +∂L∂ux

δux +∂L∂uxx

δuxx dtdx

=

∫ ∫ (∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

+d2

dx2

∂L∂uxx

)δu dtdx

+boundary terms (190)

To get the middle two terms I integrated by parts once, and to get the lastterm I integrated by parts twice; i.e. in other words I use

d2

dx2

(∂L∂uxx

δu

)− 2

d

dx

(∂L∂uxx

δux

)= δu

d2

dx2

(∂L∂uxx

)− δuxx

(∂L∂uxx

)(191)

since everything on the LHS of the first line is a total derivative so again givingboundary terms that are irrelevant to the E-L equations. These are as beforewith the extra uxx term;

boundary terms =

∫ ∫d

dx

(∂L∂ux

δu+ 2∂L∂uxx

δux −d

dx

(∂L∂uxx

δu

))+d

dt

(∂L∂ut

δu

)dxdt∫ ∫

d

dx

(∂L∂ux

δu+∂L∂uxx

δux −d

dx

(∂L∂uxx

)δu

)+d

dt

(∂L∂ut

δu

)dxdt. (192)

The E-L equation becomes

∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

+d2

dx2

∂L∂uxx

= 0 (193)

With the choice of lagrangian we get (using w in place of u) that

0 = − d

dt(wx2

)− d

dx(3w2

x +1

2wt) +

d2

dx2wxx

= −wxt − 6wxwxx − wxxxx (194)

so wx obeys the KdV equation!

55

10.2 Additional Lagrangians 2: The non-linear Schrodinger(NLS) equation

The BT for this equation was done in question 3 of sheet 2. Consider

L =i

2(u∗ut − uu∗t )− |ux|2 −

λ

2|u|4. (195)

u here is complex but S is still real since L ∗ = L . Treat u and u∗ as indepen-dent (Like z and z∗) when differentiating. Varying by u∗ we get

0 =∂L∂u∗

− d

dt

∂L∂u∗t

− d

dx

∂L∂u∗x

= − i2ut − uxx + λ|u|2u. (196)

Varying by u would just give the complex conjugate of this, so doesn’t addany new information. This is a version of the NLS equation. It’s an importantalso integrable equation with applications to fibre optics.

56

10.3 Noether’s theorem: conserved quantities from sym-metries of the Lagrangian

Return to conservation laws. Recall we seek Xand T such that

Xx + Tt = 0

X|x=±∞ = 0 (197)

so that dt∫Tdx =

∫Ttdx = −

∫Xxdx = 0.

10.3.1 Time translation symmetry

Now consider the infinitessimally small constant shift t→ t+ε. If S is invariantunder this shift it is called a time-translation symmetry of the action, and thereis a conserved current associated with that given by the boundary terms. Wesee it as follows. Under the shift we have by Taylor expanding that

t → t+ ε

u(x, t) → u(x, t+ ε) = u(x, t) + εut

ut(x, t) → ut(x, t+ ε) = ut(x, t) + εutt

ux(x, t) → ux(x, t+ ε) = ux(x, t) + εuxt

uxx(x, t) → uxx(x, t+ ε) = uxx(x, t) + εuxxt

...etc (198)

The extra bits on the RHS we can take to be δu, δut, δux, δuxx etc. Now lookat the shift in the action, or rather everything inside the integral. Since theE-L equations are satisfied locally, all we have left are the boundary terms.These are (assuming for the moment that L = L (u, ut, ux) only and does notdepend on uxx or very importantly explicitly on x or t)

δL =d

dx

(∂L∂ux

δu

)+d

dt

(∂L∂ut

δu

)= ε

d

dx

(∂L∂ux

ut

)+ ε

d

dt

(∂L∂ut

ut

)(199)

dividing by ε and taking the ε→ 0 limit to get δLε→ dL

dtwe have the relation

d

dx

(∂L∂ux

ut

)+d

dt

(∂L∂ut

ut − L)

= 0 (200)

This relation is in precisely the form Xx + Tt = 0. When the action doesdepend on uxx as well, we get the slightly more complicated

d

dx

(∂L∂ux

ut +∂L∂uxx

uxt −d

dx

(∂L∂uxx

)ut

)+d

dt

(∂L∂ut

ut − L)

= 0 (201)

57

In either case the conserved current is

H =∂L∂ut

ut − L (202)

Some of you may recognize this as the Hamiltonian (with p ≡ ∂L /∂ut) hencethe name H. So the hamiltonian is a conserved current because the actionis time-translation invariant. Conversely an explicit time dependence in theLagrangian would break time translation invariance, and the current would nolonger be conserved. This is a very deep relation that has many consequences.

10.3.2 Examples

SG:

H =∂L∂ut

ut − L = u2t −

u2t

2+u2x

2+ (1− cosu)

=u2t

2+u2x

2+ (1− cosu)

KdV: Here we recognize the charge Q3 as the hamiltonian! This is like theenergy of the water wave;

H =∂L∂wt

wt − L =1

2wxwt −

1

2wxwt − w3

x +1

2w2xx

= −w3x +

1

2w2xx

= −u3 +1

2u2x.

NLS: Here the answer is again associated with energy

H =∂L∂ut

ut +∂L∂u∗t

u∗t − L = uti

2u∗ − u∗t

i

2u− i

2(u∗ut + uu∗t ) + |ux|2 +

λ

2|u|4.

= |ux|2 +λ

2|u|4.

10.3.3 Space translation symmetry

Now consider the infinitessimally small constant shift x→ x+ε. If S is invari-ant under this shift it is called a space-translation symmetry of the action, andagain there is a conserved current associated with that given by the boundaryterms. Under the shift we have

x → x+ ε

u(x, t) → u(x, t+ ε) = u(x, t) + εux

ut(x, t) → ut(x, t+ ε) = ut(x, t) + εutx

ux(x, t) → ux(x, t+ ε) = ux(x, t) + εuxx

uxx(x, t) → uxx(x, t+ ε) = uxx(x, t) + εuxxx

...etc (203)

58

To get the conserved current we can read of from eq.201 that

d

dx

(∂L∂ux

ux +∂L∂uxx

uxx −d

dx

(∂L∂uxx

)ux − L

)+d

dt

(∂L∂ut

ux

)= 0 (204)

The only difference from the time-translation case is that the −L went in thed/dx term. In this case the conserved current is

P =∂L∂ut

ux (205)

10.3.4 Examples

Sine Gordon: P = ∂L∂utux = utux

This is the momentum density

KdV: P = ∂L∂wt

wx = 12w2x = 1

2u2

We recognize the charge Q2 as the momentum!

NLS: P = ∂L∂utux + ∂L

∂u∗tu∗x = i

2(uxu

∗ − u∗xu) = i2∂x|u|2

Again this is linked to momentum density in QM. (c.f. p = i∂x and〈p〉 =

∫u∗pu)

10.4 Other conserved currents beyond energy and mo-mentum

The currents derived above are the standard ones of many theories. Butwhere does Q1 =

∫u dx come from? Consider instead shifts in w. The KdV

lagrangian has an obvious symmetry when w → w + ε where ε is again aconstant.

We will to maintain generality work with a generic lagrangian whose fieldis named u and assume that ε is not necessarily a constant, but can also bea function of u, so ε = ε(u). Now we have simply that δu = ε, but δux = εxand δut = εt. We get

d

dx

(∂L∂ux

ε+∂L∂uxx

εx −d

dx

(∂L∂uxx

)ε

)+d

dt

(∂L∂ut

ε

)= 0 (206)

The ε∂L /∂u differential term in the limit goes into the E-L equations and wehave left the ε inside the derivatives now, as it is not necessarily (although isfor KdV) constant. The conserved current is

T =∂L∂ut

ε (207)

KdV: We see immediately (renaming u → w) that for the symmetry w →w + a (a =constant)

T =a

2wx =

a

2u. (208)

59

So the charge Q1 is associated with symmetry under constant shifts inw.

NLS has a symmetry under the global rotation u→ eiθu and simultaneouslyu∗ → e−iθu∗ where θ is constant. Here for infinitessimal θ we haveε = iθu and also a shift ε∗ = −iθu∗ for u∗. We find

T =∂L∂ut

ε+∂L∂u∗t

ε∗

= −θ2(u∗u− uu∗) = −θ|u|2 (209)

This conservation law gives∫|u|2dx = 0, i.e. conservation of total prob-

ability.

Generally any continuous symmetry gives a conservation law via Noether’stheorem. In more comlicated theories the set of conserved currents form agroup called a Lie group. e.g. in special relativity the energy and momentumcombine to form the (conserved) energy momentum tensor. The conservedgroup is the Poincare group. Principle underlying all modern day particlephysics. Unfortunately all the other conserved currents of KdV are difficult tounderstand this way (current problem).

60

11 Hirota’s Method

It’s remarkable how many methods for solving PDEs the KdV equation hasstimulated. In this section we look at another one due to Hirota in 1971. Wehave already seen that a linear equation of the form

∂nxu− a ∂mt u = 0 (210)

where a is a constant, can easily be solved. E.g. we can use the linear super-position principle and write

u(x, t) =

∫ ∞

−∞f(k)ei(kx−ωt)dk. (211)

Applying the differential equation we get

(ik)n − a(iω)m = 0. (212)

A simple single wave solution is

u = ei(kx−ωt) (213)

whereω = −i ((ik)n/a)

1m . (214)

Hirota noticed that there is a moderately simple way to get solutions forbilinear PDEs; that is equations of the form

ffx..x + ft..tfx..x + fxtf + ... (215)

i.e. all terms can have any number of x, t differentiations acting on always aproduct two f ’s. It involves introducing a new operator (Hirota’s operators)with their own algebra. The technique is very eneral and often gives solutionswhere previously none where available (e.g. recently here in Durham). Beforewe do so lets first look at some hints at how to get simple KdV solutions.

11.1 Hint 1: Burger’s equation

Recall from question 6 on sheet 1, that equations of the form

ut + uux + λuxx = 0

are converted into linear equations by the transformation

u = 2λ(log f)x.

Comparison with the linear part of the KdV equation (that is ut + uxxx) sug-gests that it might be worth trying a substitution with λ = 1;

u = 2(log f)xx

61

11.2 Hint 2: The one soliton solution

Now recall from last term that via the BT we found a solution u = wx where

w = 2√λ tanh(

√λ(x− x0 − 4λt)) (216)

Call X = x− x0 − 4λt.Then

w = 2√λe2√λX − 1

e2√λX + 1

= 2√λ

2e2√λX − (1 + e2

√λX)

e2√λX + 1

= 4√λ

e2√λX

e2√λX + 1

− 2√λ

= 2∂x log(e2√λX + 1)− 2

√λ (217)

Since we also know w → w+ const gives the same solution we could have used

w = 2∂x log(e2√λX + 1) (218)

This is of precisely the form guessed in hint 1, with f ≡ e2√λX + 1.

11.3 The KdV equation in Bilinear form

Motivated by this let’s try substituting w = 2(log f)x = 2fx/f into the KdVequation. First write u = wxand integrate once to get a slightly different p.d.e.for w;

A(t) =

∫(wxt + 6wxwxx + wxxxx)dx

= wt + 3w2x + wxxx (219)

whereA(t) is a t-dependent constant of integration. Now consider the boundaryconditions to determine A(t). At x → ±∞ we look for solutions that havewt = wx = wxxx = 0∀t so we can take A(t) = 0 and consider

wt + 3w2x + wxxx = 0 (220)

Now get some identities first by differentiating;

w

2=

fxf

wt2

=fxtf− fxft

f 2

wx2

=fxxf− f 2

x

f 2

wxx2

=fxxxf

− 3fxfxxf 2

+2f 3

x

f 3

wxxx2

=fxxxxf

− 4fxfxxxf 2

− 3f 2xx

f 2+

12f 2xfxxf 3

− 6f 4x

f 4(221)

62

(Notice that you always have the same numbers of f s on the top as bottomwhen you diffentiate logs.) So now

0 = wt + 3w2x + wxxx =

2fxtf

− 2fxftf 2

+ 12

(fxxf− f 2

x

f 2

)2

+2fxxxxf

− 8fxfxxxf 2

− 6f 2xx

f 2+

24f 2xfxxf 3

− 12f 4x

f 4

=2fxtf

− 2fxftf 2

+2fxxxxf

− 8fxfxxxf 2

+6f 2

xx

f 2(222)

orffxt − fxft + ffxxxx − 4fxfxxx + 3f 2

xx = 0 (223)

This is the bilinear form of the KdV equation. Despite the fact that somenontrivial cancellations took place it looks more complicated than the originalproblem, however its special form makes it possible to solve using bilinearoperators.

11.4 Hirota’s Bilinear operators D(f.g).

First note that the first two terms of eq.223 are similar to

1

2∂x∂tf

2 = ffxt + fxft.

The only difference is the sign. Hirota’s operators make it possible to writethe opposite sign term in a similarly simple way. They map a pair of functionsto one function as follows; For m,n ≥ 0

Dmt D

nx(f.g) =

(∂

∂t− ∂

∂t′

)m (∂

∂x− ∂

∂x′

)n

f(x, t)g(x′, t′)

∣∣∣∣x′=x, t′=t

(224)

The usefulness of this operator becomes obvious when you do some examples;

Example 1:

Dt(f.g) =

(∂

∂t− ∂

∂t′

)f(x, t)g(x′, t′)

∣∣∣∣x′=x, t′=t

= ft(x, t)g(x′, t′)− f(x, t)gt′(x

′, t′)|x′=x, t′=t= ftg − fgt (225)

all functions of x, t. In particular this means that

Danything(f.f) = 0 (226)

63

Example 2:

DtDx(f.g) =

(∂

∂t− ∂

∂t′

) (∂

∂x− ∂

∂x′

)f(x, t)g(x′, t′)

∣∣∣∣x′=x, t′=t

=

(∂

∂t− ∂

∂t′

)(fx(x, t)g(x

′, t′)− f(x, t)gx′(x′, t′))

∣∣∣∣x′=x, t′=t

= fxtg − fxgt − ftgx + fgxt (227)

all functions of x, t. In particular this means that

1

2DtDx(f.f) = ffxt − fxft (228)

i.e. the opposite sign combination appearing in eq.223.

Example 3:

D2x(f.g) =

(∂

∂x− ∂

∂x′

)2

f(x, t)g(x′, t′)

∣∣∣∣∣x′=x, t′=t

=

(∂

∂x− ∂

∂x′

)(fx(x, t)g(x

′, t′)− f(x, t)gx′(x′, t′))

∣∣∣∣x′=x, t′=t

= fxxg − 2fxgx + fgxx (229)

all functions of x, t. Again this looks a bit like the differentiation of aproduct. But be careful because

D2x(f.f) = 2ffxx − 2f 2

x (230)

even though Dx(f.f) = 0, so importantly the operators are not associa-tive since D2

x(f.f) 6= Dx(Dx(f.f)). For starters the RHS is meaninglessbecause the two functions are already mapped by the first Dx to onefunction, so the second Dx doesn’t have two functions to “eat”.

Example 4:

D4x(f.g) =

(∂

∂x− ∂

∂x′

)4

f(x, t)g(x′, t′)

∣∣∣∣∣x′=x, t′=t

=

(∂

∂x− ∂

∂x′

)3

(fx(x, t)g(x′, t′)− f(x, t)gx′(x

′, t′))

∣∣∣∣∣x′=x, t′=t

=

(∂

∂x− ∂

∂x′

)2

(fxxg − 2fxgx′ + fgx′x′)

∣∣∣∣∣x′=x, t′=t

=

(∂

∂x− ∂

∂x′

)(fxxxg − 3fxxgx′ + 3fxgx′x′ − fgx′x′x′)

∣∣∣∣x′=x, t′=t

= (fxxxxg − 4fxxxgx′ + 6fxxgx′x′ − 4fxgx′x′x′ + fgx′x′x′x′)|x′=x, t′=t= fxxxxg − 4fxxxgx + 6fxxgxx − 4fxgxxx + fgxxxx (231)

64

all functions of x, t. Again it looks just like a differentiated product of fand g but with sign changes on the odd terms. In particular

D4x(f.f) = 2fxxxxf − 8fxxxfx + 6f 2

xx (232)

Comparing examples 2 and 4 we see that eq.223 can be written in a very simpleway;

(DtDx +D4x)(f.f) = 0 (233)

11.5 The 1-soliton solution

In Hirota form the multisoliton solutions are sums of exponential of linearexpressions in x and t. For the one soliton solution try

f = e0 + eθ (234)

where θ = ax+ bt+ c.

Lemma 1: A useful result is

Dmt D

nx(e

θ1 .eθ2) = (b1 − b2)m(a1 − a2)

neθ1+θ2 (235)

Again this looks like the differentiation of products with the oppositesigns.

Proof: by induction. Let

Dmt D

nx(e

θ1 .eθ2) = (b1 − b2)m(a1 − a2)

neθ1+θ2

= ((b1 − b2)m(a1 − a2)

neθ1(x,t)+θ2(x′,t′))∣∣∣x′=x, t′=t

Then

Dmt D

n+1x (eθ1 .eθ2) = (b1 − b2)

m(a1 − a2)n

(∂

∂x− ∂

∂x′

)eθ1(x,t)+θ2(x′,t′))

∣∣∣∣x′=x, t′=t

= (b1 − b2)m(a1 − a2)

n (a1 − a2)eθ1(x,t)+θ2(x′,t′))

∣∣∣x′=x, t′=t

= (b1 − b2)m(a1 − a2)

n+1eθ1+θ2 (236)

and similarly for Dt operator. Finally the equation is obviously true forn = 1, m = 0 and n = 0, m = 1 so by induction must be true for alln,m.

In particular we find

Dmt D

nx(e

θ.eθ) = 0 (237)

Dmt D

nx(e

θ.1) = bmaneθ (238)

Dmt D

nx(1.e

θ) = (−1)n+mbmaneθ (239)

65

Then KdV on f = 1 + eθ gives

0 = (DtDx +D4x)(1 + eθ.1 + eθ) = (DtDx +D4

x)[(1.1) + (1.eθ) + (eθ.1) + (eθ.eθ)]

= (DtDx +D4x)[(1.e

θ) + (eθ.1)]

= 2ba eθ + 2a4eθ (240)

so that a solution requires a(b + a3) = 0. a = 0 would give no x-dependenceand hence (by the boundary conditions) u = 0. So take b = −a3 and then

θ = ax− a3t+ c

and hence

u = 2∂2x(log(1 + eax−a

3t+c))

=a2

2sech 2[

1

2(ax− a3t+ c)] (241)

This is the one-soliton solution with v = a2!

11.6 Multi-soliton solutions

General idea: organize the calculation perturbatively in some parameter ε(secret hope is that the series will terminate at some power of ε to giveexact results rather than an infinite series in ε).

To begin with we put

f(x, t) = 1 +∞∑n=1

εnfn(x, t) (242)

and collect powers of ε. We start with 1 as the ε0 term as it appeared in theone-soliton solution. Defining

B = (DtDx +D4x) (243)

we want to find solutions such that

B(f.f) = 0. (244)

Expanding as for the one soliton case we have

0 = B(f.f) = B(1.1) + ε(B(1.f1) +B(f1.1)) + ε2(B(1.f2) + (B(f1.f1) + (B(f2.1))

=∞∑n=0

εnn∑

m=0

B(fn−m.fm)

so actually have to solve an infinite set of equationsn∑

m=0

B(fn−m.fm) = 0 (245)

We can think of this as a recursive set of equations to get fn from f1 . . . fn−1.That is we begin with f1 and solve iteratively to get f2,3,.. hoping that itterminates at some point. To do this we need

66

Lemma 2: Another useful result is

Dmt D

nx(f.1) = (−1)n+mDm

t Dnx(1.f) (246)

Proof: for this we need simply note that we can swop the primed and un-primed variables since “prime” or “unprime” are simply labels; i.e.

Dmt D

nx(f.g) =

(∂

∂t− ∂

∂t′

)m (∂

∂x− ∂

∂x′

)n

f(x, t)g(x′, t′)

∣∣∣∣x′=x, t′=t

= (−1)n+m

(∂

∂t′− ∂

∂t

)m (∂

∂x′− ∂

∂x

)n

g(x′, t′)f(x, t)

∣∣∣∣x′=x, t′=t

= (−1)n+m

(∂

∂t− ∂

∂t′

)m (∂

∂x− ∂

∂x′

)n

g(x, t)f(x′, t′)

∣∣∣∣x′=x, t′=t

= (−1)n+mDmt D

nx(g.f) (247)

where in the 3rd line I swopped x↔ x′ and t↔ t′.

With this result eq.245 becomes

∂x(∂t + ∂3x)fn = −1

2

n−1∑m=1

B(fn−m.fm) (248)

with∂x(∂t + ∂3

x)f1 = 0 (249)

Beginning with f1 we may iterate to find all the fn and note that at any orderwe are now solving linear partial differential equations. A simple particularsolution to the last equation would be

f1 = eax−a3t+c (250)

But now notice what happens for f2. By eq.248 we have

∂x(∂t + ∂3x)f2 = −1

2B(f1.f1)

= 0 (251)

where the last line follows by eq.237. Now the f3...∞ all vanish trivially. Theseries has terminated with

f = 1 + εf1 (252)

We can always absorb ε into the constant c and we recognize the single solitonsolution.

The clever bit: Our hope is to find other solutions where the series termi-nates in this way, so that the solution is exact. Consider beginning with

f1 =N∑i=1

eaix−a3i t+ci (253)

67

which clearly satisfies eq.250, since it is linear and we can simply addsolutions. The f2 equation now no longer vanishes by virtue of the crossterms in B(f1.f1) . However the operation of B retains the form of theexponentials which are simply multiplied together by virtue of eq.235.The N exponentials can be thought of as N seperate variables in a setof simultaneous equations since they persist at every iteration. Sincethere are only N of them we can have at most have N equations beforethe system is overdetermined, and so the series must terminate after Nterms. This is the N soliton solution!

11.6.1 Two soliton example

We have already seen the single soliton solution. Let’s verify that we recoverthe two soliton solution. Take

f1 = eθ1 + eθ2 (254)

where θ1,2 are as above chosen to satisfy the f1 equation

θi = aix− a3i t+ ci (255)

The f2 equation becomes

∂x(∂t + ∂3x)f2 = −1

2B(f1.f1)

= −1

2B(eθ1 .eθ2)− 1

2B(eθ2 .eθ1)

= −(a1 − a2)((b1 − b2) + (a1 − a2)3)eθ1+θ2

= −(a1 − a2)(−(a31 − a3

2) + (a1 − a2)3)eθ1+θ2

= (a1 − a2)2((a2

1 + a1a2 + a22 − a2

1 + 2a1a2 − a22)e

θ1+θ2

= 3a1a2(a1 − a2)2eθ1+θ2 (256)

Since the equation is linear in f2 it has an obvious solution of the form

f2 = Aeθ1+θ2 .

so we just have to determine A. Substituting into the f2 equation we get

A(a1 + a2)(−(a31 + a3

2) + (a1 + a2)3) = 3a1a2(a1 − a2)

2

A 3a1a2(a1 + a2)2 = 3a1a2(a1 − a2)

2

or

A =(a1 − a2)

2

(a1 + a2)2. (257)

You can now check (exercise) that the RHS of the f3 equation vanishes andthe series terminates. The solution is then

f = 1 + εeθ1 + εeθ2 +(a1 − a2)

2

(a1 + a2)2ε2eθ1+θ2 (258)

68

Again we can absorb the ε into the constants c1,2(or equivalently set ε = 1).Finally we just need to differentiate to get the original w (or u) using

w = 2fxf

= 2a1e

θ1 + a2eθ2 + (a1−a2)2

(a1+a2)eθ1+θ2

1 + eθ1 + eθ2 + (a1−a2)2

(a1+a2)2eθ1+θ2

It is easier in this format to study the asymptotics of the two soliton solution.As in the BT discussion, first choose a coordinate system with origin at thecentre of the around i = 1 or 2 soliton; take t→ ti+ δt with δx = x−x0− vitiwhere vi = a2

i and let ti → ±∞. We have

θi =

√vi2

(δx− viδt)

θj 6=i =

√vj

2(δx− vjδt) +

√vj

2(vi − vj)ti (259)

Near the i soliton we have (by definition) δx, δt ≈ 0, so that θi is small butθj 6=i → ±∞ as follows;

t→ −∞ t→ +∞i = 1 θj=2 → −∞ θj=2 → +∞i = 2 θj=1 → +∞ θj=1 → −∞

When θj 6=i → −∞ the f clearly reduces to the single soliton case, f →1 + eθi . When θj 6=i → +∞ we have to do a little more work. To leading orderwe have

f = eθj

(1 +

(a1 − a2)2

(a1 + a2)2eθi + . . .

)(260)

where the ellipsis indicates terms of order e−θ2 . But then

w = 2∂x log f

= 2aj + 2∂x log

(1 +

(a1 − a2)2

(a1 + a2)2eθi

). (261)

Since w → w + const gives equivalent solutions we may ignore the constantpiece. The remaining solution is equivalent to an f of the form

f ≡(

1 +(a1 − a2)

2

(a1 + a2)2eθi

). (262)

This is the “original” single i-soliton solution but with a phase shift of

δx0 = − 2√vi

log

(|√v1 −

√v2|√

v1 +√v2

), (263)

precisely what we found for the 2 KdV soliton solution using the much morecomplicated BT method.

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11.6.2 N-soliton solution

Finally I’ll just present the general solution which may be proved by induction.(This is a very difficult exercise.) Define an N ×N matrix SN with elements

(SN)ij = δij +2aie

θi

ai + aj(264)

Then the N -soliton solution is

u = 2∂2x log f (265)

wheref = det(SN).

You can easily check the single and two-soliton solutions.

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12 Overview for the Inverse Scattering Method

12.1 Initial value problems

So far we have asked the question whether for a particular p.d.e. it is possibleto construct particular solutions. We have mainly been interested in singleor multi-soliton solutions. However you can also ask whether you can find ageneral solution, as we were able to for the Liouville equation. The questionis

If we are given a wave equation and sufficient initial data at t = 0(i.e. the functions u(x, 0), ut(x, 0), . . .), can we find u(x, t) for all t?

This is known as an initial value problem. First for there to be a uniquesolution what initial data do we need?

• If eqn. is 1st order in t (e.g. KdV equation) then we need u(x, 0)

• If eqn. is 2nd order in t (e.g. SG equation) then we need u(x, 0), ut(x, 0)

Why? Because we can use the equation itself to solve for higher t derivatives.e.g. in the KdV case, if you know u(x, 0) you can solve the KdV equation toget ut(x, 0).

So far, we might be able to spot a particular solution and (assuming unique-ness) from the initial form, but in general this is not possible. e.g. considerthe three initial functions

1. u(x, 0) = 2sech 2x

2. u(x, 0) = 2.01sech 2x

3. u(x, 0) = 6sech 2x

The first case can be spotted as a t = 0 snapshot of a one-soliton solution withv = 4;u(x, t) = 2sech 2(x−4t) but the other two are not so simple. Case 2 turnsout to be a soliton plus oscillatory “junk” whereas case 3 is actually a snapshotof a two soliton solution. The evolution can of course be done numerically, asbelow for these three cases, but we want to understand it analytically.

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The remainder of the course will show how this can be done and discussessome interesting side issues that have been stimulated by this question; forthe moment I’ll give an overview of the technique that you should keep at theback of your minds.

12.2 The linear initial value problem

It helps to have an analogous situation to guide us, and the obvious one isthe use of Fourier Transforms for solving linear problems such as the Heat(diffusion) equation and the Klein-Gordon equation. We have already seenthe former in lectures so let’s look at the latter. This is exactly the sameprocedure as for the diffusion equation, but the point of going through thisagain is to highlight important features

utt − uxx +m2u = 0 (266)

where I have set v = 1. As it’s second order, the initial conditions we need areu and ut. Let’s define

A(x) = u(x, 0)

B(x) = ut(x, 0) (267)

72

As discussed previously we can solve this by using a Fourier transform in thex−coordinate. Let

u(x, t) =

∫ ∞

−∞u(k, t)eikxdk

u(k, t) =1

2π

∫ ∞

−∞u(x, t)e−ikxdx

Note that u(k, t) is a function of t, and (to make connection with the analysisfor the KdV equation later) I’ll call this the scattering data; it tells us at anytime what sinusoidal waves u is composed of. Now applying the KG equationto u as written above gives

utt + (k2 +m2)u = 0 (268)

This is an equation that can be simply solved to give the time-dependence ofthe scattering data;

u = f(k) eiωt + g(k) e−iωt, (269)

where f and g are to be determined by the initial conditions, and

ω =√k2 +m2 (270)

is the dispersion relation.We can now determine the “spectrum” of positive and negative modes f

and g from the FT of the initial conditions. When t = 0 we have

A = f + g

B = iω(f − g) (271)

so that

f =iωA+ B

2iω

g =iωA− B

2iω(272)

where

A(k) =1

2π

∫ ∞

−∞u(x, 0)e−ikxdx

B(k) =1

2π

∫ ∞

−∞ut(x, 0)e−ikxdx. (273)

Finally our solution can be written as

u(x, t) =

∫ ∞

−∞u(k, t)eikxdk

=

∫ ∞

−∞

(iωA+ B)

2iωei(kx+ωt) +

(iωA− B)

2iωei(kx−ωt) dk

=

∫ ∞

−∞(A cosωt+

B

ωsinωt ) eikx dk

73

An interesting thing just happened which has a direct analogy in the KdVequation. The reason we were able to find the solution is that the energyspectrum of positive and negative modes was time-independent (i.e. u+ = |f |2and u− = |g|2 are independent of t). If we break the solution down into threesteps,

1. Perform the FT in the x coordinate and determine the initial “scatteringdata” u(k, 0) from the initial conditions u(x, 0), ut(x, 0).

2. Evolve u(k, t) forwards in time using the evolution (determined by theKG equation) in eq.269

3. Perform the inverse FT back to get u(x, t),

it is the second of these steps that was allowed by the time-independent spec-trum property. Otherwise the time evolution would have mixed everything upand it would have been very difficult to get the final spectrum from the initialone (i.e. the FT would have been of no use!). It can be represented as ...

12.3 Outline of the method, and the KdV-Schroedingerconnection

The method for the KdV equation was discovered by Gardner, Greene, Kruskaland Miura in the late 60’s and is analogous. At first sight the constancy-of-the-spectrum property looks like a unique property of linear equations, butlinear equations are just the simplest of a much wider class of equations calledthe Lax KdV hierarchy (that we’ll get to much later) where this type of be-haviour occurs. In particular something very similar (although much morecomplicated) can be done for the KdV equation. Let’s generalize the proce-dure above to the KdV equation;

74

The method now relies on a connection between KdV and the Schroedingerequation. The equation we ultimately wish to solve is the KdV equation

ut + 6uux + uxxx = 0 (274)

Recall the generalized Miura transformation

u = λ− v2 − vx (275)

where λ is constant; if v satisfies

vt + 6(λ− v2)vx + vxxx = 0

it implies that u given by the above is a solution of the KdV equation. Nowthink about it the other way round; if u is a known solution of the KdVequation we can solve eq.275 to find v; writing

v =ψxψ

(276)

equation 275 becomes

u = λ− ψxxψ

i.e.ψxx + uψ = λψ (277)

This is remarkable because it is the time-independent Schroedinger equationfor the Sturm-Liouville problem (time appears only as a parameter in u(x, t))with eigenvalue λ(t) (since u is a time dependent potential we generally expectλ to be time dependent as well). The QM equation for a particle in a potential−u. A great deal was known at the time about such equations (otherwise theconnection with QM is unimportant).

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12.4 The recipe;

The general instructions for constructing the time dependent solutions go asfollows

Disassembly (Scattering): Note that t is only a parameter in eq.277. Beginwith initial data u(x, 0) which plays the role of the potential. Now foreach λ, the ψ is a different eigenfunction. Each eigenfunction describesthe scattering of a particle off the potential with certain reflection andtransmission coefficients (i.e. the asymptotic values of ψ at x → ±∞)- we collect these coefficients which are our initial scattering data. Hereλ is playing the role of k in the FT, and effectively the FT has beenreplaced with determining the components of v = ψx/ψ from u

Time evolution: Now we have to evolve the eigenvalues forward in time.There is an amazing fact that helps us now - for potentials u that obeythe KdV equation we will be able to show that actually any spectrum ofλ is time independent! The spectrum is a constant and all we need todo is evolve the scattering data in a very simple manner.

Reassembly (Inverse scattering): The final step is to reconstruct the po-tential u(x, t) at time t by the inverse scattering. That is, given the setof scattering data can have to reconstruct the potential that the particleshave scattered off. (c.f. “can you hear the shape of a drum”). That thisis possible for the Sturm Liouville equation was already known.

Each of these steps is rather lengthy so I advise you to keep referring back tothis overview as we go along. It will take a little time to get straight what ishappening!

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13 Disassembly: Scattering theory

We now begin some lengthy detours to treat each of the necessary ingredients,beginning with scattering theory. We want to analyse the possible solutionsto

(∂2x + u)ψ = λψ (278)

with ψ remaining bounded ∀x. For shorthand I will call the operator on theLHS L so that

Lψ = λψ. (279)

Here u is considered to be at some fixed time t. This eqn is the time indepen-dent Schroedinger eqn.

13.1 Physical interpretation

We begin with the time dependent Schroedinger equation from QM

(−∂2x + V (x))Ψ(x, τ) = i∂τΨ(x, τ) (280)

(Here I use τ for time because it has nothing to do with the time in the KdVequation or in u(x, t). That time we are considering as fixed.) This describesa particle moving in a potential V (this is simply a fact - but to motivate itwe replace the operators i∂x → p/

√2m and i∂τ → E and find an equation for

the non-relativistic energy, E ≡ p2/2m + V ). |Ψ(x)|2becomes the probabilitydensity for finding the particle at x.

To solve this eqn we can separate variables

Ψ(x, t) = ψ(x)T (τ)

iT

T=

(V ψ − ψ′′)

ψ= const = k2 ≡ −λ

where k2 is a constant. Then T (τ) = e−ik2τ and

ψ′′ − V ψ = k2ψ.

Hence we see the interpretation of −λ as the energy eigenvalue and u as thenegative potential. If limx→±∞(V (x)) = 0 and k2 > 0 then the x dependentfactor asymptotically goes like

ψ(x) → e±ikx

and the full solution looks like

Ψ(x, t) = e±ikx−ik2τ

which is a plane wave either going left (−ikx) or right (+ikx). Inbetweenthe solution ψ(x) will be doing something horrible depending on V , but thecoefficients of asymptotic left and right moving waves can be relatively easilydetermined and these will be the all important scattering data.

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Example 1: V (x) = 0

This case includes the asymptotic solution above to the equation

ψ′′ = k2ψ (281)

which has 2 casesa) k2 > 0;we have already seen that ψ = Aeikx + Be−ikx is a right or

left-moving wave which is always bounded.b) k2 =< 0; This would be negative energy. Call k2 = −µ2 and then

ψ = Aeµx + Be−µx. The only way to keep the solution bounded at x→∞ isto have A = B = 0 so that there are no appropriate solutions of this kind.

Conclusion: if u = 0 then Lψ = λψ has a solution for all λ ≤ 0 (positiveenergy) and no bounded solutions for λ > 0 (negative energy). The energyspectrum of solutions looks like

Example 2: V (x) = aδ(x) where a = const (the Dirac delta function)

This situation is particles penetrating an infinitely thin and infinititely highbarrier of area under the curve =1. See the Fourier sheet for a definition of δ,but for this discussion we can use the formal definition of δ(x) as a functionthat obeys

f(x) =

∫f(x′)δ(x− x′)dx′ (282)

and is exactly zero for x 6= 0.a) k2 > 0;we have already seen that ψ(x) = Aeikx + Be−ikx is a right or

left-moving wave. We can split it into two regions on either side of the spikewith different constants in each region

ψ(x < 0) = A−eikx +B−e

−ikx

ψ(x > 0) = A+eikx +B+e

−ikx (283)

78

We find the constants by matching at x = 0. It can be shown that at x = 0the function must be discontinuous (for starters a discontinuity in ψ wouldindicate an unphysical sudden change in the probability density). So we canset

A− +B− = A+ +B+. (284)

The equation now becomes

−ψ′′ + aδ(x)ψ = k2ψ (285)

Integrate over x gives

[−ψ′]ε−ε + aψ(0) = k2

∫ ∞

−∞ψdx

= k2

∫ ε

−εψdx

= 0 (286)

where ε is an infinitessimally small region near x = 0, and I used the fact thatψ is bounded in the region near x = 0 and the integrals of sinusoidal functionsover x vanishes. This equation gives

ik(A+ −B+ − A− +B−) = a(A− +B−) = a(A+ +B+) (287)

This is two equations in 4 unknowns. Solving for A− and B− we find

A− = A+(1− a

2ik) +B+

a

2ik

B− = B+(1 +a

2ik) + A+

a

2ik(288)

This gives the general solution. There are 2 unknowns since it is a 2nd ordero.d.e. for ψ(x).

Important special case: Consider a flow of particles incident from theleft. The incoming probability density is just normalized by looking at the fluxof particles so it doesn’t carry any interesting information. We can normalizeit to 1 by convention, so that there is a unit flux of particles coming in fromx = −∞ and a zero flux from x = +∞. This plus the equation above gives

A− = 1

B+ = 0

A+ =2ik

2ik − a

B− =a

2ik − a(289)

The total solution can be written

ψ =

{eikx + a

2ik−ae−ikx x < 0

2ik2ik−ae

ikx x > 0.(290)

79

The coefficient of e−ikx is called the reflection coefficient, sometimes writtenR(k) = a/(2ik − a), and the coefficient of eikx on the right of the barrier iscalled the transmission coefficient, sometimes written T (k) = 2ik/(2ik − a).The physical interpretation is clear;

• |R(k)|2 gives the probability that the particle is reflected by the barrierand |T (k)|2 is the transmission probability

• |R|2 + |T |2 = 1 since the total probability must be one

b) k2 = −µ2 < 0;for the V = 0 case we got no solutions but here we geta surprise - an acceptable (i.e. bounded) solution exists. Begin be replacingk = iµ everywhere so that the solution is

ψ =

{e−µx − a

2µ+aeµx x < 0

2µ2µ+a

e−µx x > 0.(291)

This looks unbounded, since eµx diverges for x → ∞ and e−µx diverges asx → −∞. However the equation is linear so we can certainly get a newsolution if we divide by the transmission coefficient T = 2µ/(2µ+ a) which isjust a constant - all that changes is the normalization of ψ. This gives

ψ =

{ 2µ+a2µ

e−µx − a2µeµx x < 0

e−µx x > 0.(292)

Generally this still diverges, but if it happens that

µ = −a2

and µ > 0 then the diverging pieces disappear. The eigenfunction is symmetricabout the origin, bounded and can be written as

ψ = e−a2|x|. (293)

80

Note that for ψ(x→ ±∞) = 0 so that the particle is stuck around x = 0.It is a QM bound state. Classically you would say that the particle is trappedbecause the energy (−λ = k2 = −a2/4) is negative. The energy spectrumlooks like the figure on the left below. Generally we find a spectrum thatlooks like the right figure, with a discrete (possibly infinite) set of negativeenergy bound states, with a continuous spectrum of postive energy states.

Example 3: V (x) = a sech 2(x)

Return now to the examples with which we began the whole discussion ofinverse scattering. The physics is the same but involves slightly more technical(but standard) functions. The equation we wish to solve is

−ψ′′ + a sech 2(x)ψ = k2ψ (294)

First we make the substitution η = tanhx, so that

d

dx= sech 2(x)

d

dη

= (1− η2)d

dη

We now need to collect a bit of technology; the equation we have to solve isnow

d

dη(1− η2)

dψ

dη+ (

k2

1− η2− a)ψ = 0 (295)

This is a standard equation that gives associated Legendre polynomials of thefirst kind (which I’ll define in a minute) commonly called Pn and Qn. Qn areunbounded so we neglect them and keep Pn. It is defined as

Pn(η) =1

n!2ndn

dηn(η2 − 1)n. (296)

81

To remove the mystery a bit the first few functions are

P1 = η

P2 = −1

2+

3η2

2

P3 = −3η

2+

5η3

2.

The associated Legendre polynomials are defined as

Pmn = (−1)m(1− η2)

m2dmPndηm

(297)

When m is non-integer or even complex, the latter can be defined in terms ofthe hypergeometric function 2F1 (you have to be a bit careful about regionsof convergence here as there are various definitions);

Γ(1−m)Pmn =

(1 + η)m2

(1− η)m2

2F1(−n, n+ 1, 1−m;1− η

2) (298)

where 2F1 has the series expansion

2F1(a, b, c; z) =∞∑k=0

Γ(k + a)Γ(k + b)Γ(c)

Γ(a)Γ(b)Γ(k + c)

zk

k!

Γ(z) =

∫ ∞

0

yz−1e−zdy. (299)

That is all the technology we need apart from the fact that for integer N ,Γ(N) = (N − 1)!. The solution is found to be (e.g. plug the differentialequation into Maple)

ψ = AP ikν (300)

where

ν =

√1− 4a

2− 1

2(301)

a) k2 > 0; In this case we expect a continuous spectrum of solutions. To getthe reflection and transmission coefficients consider first the limit x → +∞.Then η → ±1and the hypergeometric function gives 1. On the other hand wehave to use 1− η ∼ 2e−2x and then find just the transmitted wave

ψ ∼ A

Γ(1− ik)eikx (302)

In the limit x→ −∞ we have η → −1 and so (1 + η) ∼ 2e2x and in this limitwe find

ψ ∼ A(1 + η)

m2

(1− η)m2

(2mΓ(m)(1 + η)−m

Γ(−n)Γ(1 + n)+

Γ(−m)

Γ(−m− n)Γ(1−m+ n)

)∼ A

(Γ(ik)

Γ(−ν)Γ(1 + ν)e−ikx +

Γ(−ik)Γ(−ik − ν)Γ(1− ik + ν)

eikx)

(303)

82

Again the solution represents a wave coming in from infinity, except it mustbe travelling to the right. Part of the wave is reflected and part the incidentwave. We can choose A so that instead the incoming wave has unit flux; then

R(k) =Γ(−ik − ν)Γ(1− ik + ν)Γ(ik)

Γ(−ν)Γ(1 + ν)Γ(−ik)

T (k) =(−ik − ν)Γ(1− ik + ν)

Γ(1− ik)Γ(−ik). (304)

In order to simplify this expression we can use the identity Γ(z)Γ(1 − z) =π/ sin(πz). This gives

R(k) =sin(πν)

π

Γ(−ik − ν)Γ(1− ik + ν)Γ(ik)

Γ(−ik). (305)

T (k) =

√sin(π(ν + ik)) sin(π(ν − ik))

sinh(πk)

After a bit of work you can verify that |R|2 + |T |2 = 1. Interestingly R = 0when

ν = N (306)

where N is any integer, or

a = −N(N + 1). (307)

In this case the potential is called reflectionless.b) k2 < 0; Again we can put µ = −ik and we will assume that µ > 0. In

this case we have asymptotically unbounded solutions going like e±µx unlesswe can kill one of them as we did for the δ function bound state. Repeatingthat procedure, we multiply the solutions by 1/T (k) so that we have

ψ =

{sin(πν)sin(πµ)

eµx + Γ(1+µ)Γ(µ)Γ(µ−ν)Γ(1+µ+ν)

e−µx x→ −∞e−µx x→ +∞

(308)

Now only the second piece of the x→ −∞ region of the solution is unbounded.This coefficient vanishes where the Γ-functions have poles, at µ − ν = −M(where M is a positive integer). In the case

a = −N(N + 1)

we have µ = N −M giving N discrete eigenvalues corresponding to the Nsolitons in the solution, and there are no non-trivial positive energy waves.(Note asymptotically these look like the delta function.) In more general cases(e.g. u(x, 0) = 4sech 2(x)) there are both oscillatory reflections and negativeenergy bound state eigenfunctions. Thus there is a pure bound state solutiononly for the special values of a, with the solutions being of the form ψ = AP µ

N

where µ = 1... N .

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Note that we can interpret the wave as naturally splitting into solitons plusoscillatory parts. The discrete eigenvalues correspond to the soliton compo-nent, and the continuous ones are the oscillatory (dispersive) component. E.g.a = −6 corresponds to the pure 2 soliton case. (You might guess at this pointthat u(x, 0) = N(N + 1) sech 2(x) is a snapshot of a pure N soliton solutionwhich will proceed to fly off if we let it evolve in time with no oscillatorycomponent.) Naturally you expect no oscillations at infinity for such valuesand the potential is then reflectionless. (i.e. any oscillations coming in frominfinity are just “on top of” the soliton component). On the other hand whena does not take one of these values we have a reflected wave going off to infinityin one direction.

84

14 Evolving the scattering data

14.1 What data do we need?

We have found that any particular u(x, 0) leads to a set of possible eigenfunc-tions with a continuous positive spectrum and a discrete negative energy one.However it would be tedious if we always had to determine all the eigenfunc-tions for ψ in a general u. Fortunately we only need a limited set of informationabout the asymptotic behaviour of the eigenfunctions.

If the asymptotic behaviour of the discrete eigenfunctions (normalized sothat

∫|ψ|2dx = 1) is

ψn ∼ cne−µnx x→∞ (309)

with λ = −µ2n for each individual eigenvalue, and the asymptotic behaviour of

the discrete eigenfunctions (normalized so that the incoming flux = unity) is

ψ ∼{eikx +R(k)e−ikx x→ −∞T (k)eikx x→ +∞ (310)

then the collection of scattering data

{R(k), cn, µn} (311)

is completely determined by u. Conversely (and this is the importantthing) u can be reconstructed entirely from these scattering data.Consequently just knowing how these data evolve in time, enables us to re-construct u at any later time (which is of course our eventual goal). Thereason this procedure works at all is our claim is that the eigenvalues λ re-mains constant (if u is a solution of the KdV equation) which we shall verifyshortly.

The subsequent reconstruction of u is then equivalent to an inverse scatter-ing problem that we have to tackle later at the “reassembly” stage. (For thatstage, you should have in mind throwing particles at a potential and seeingwhat data you need to measure to reconstruct the entire potential - “can youhear the shape of a drum?”)

14.2 Examples

To give an idea of how to collect this date consider the examples we have seen

The delta function: Here we have R = a2ik−a . There is a single discrete

eigenvalue with ψ = Ae−a2|x| where A is a constant. Conventionally we

normalize the discrete eigenvalues such that∫|ψ|2dx = 1 so we need

A =√a/2. Consequently ψ ∼

√a2e−

a2x and so

µ1(0) = c1(0) =

√a

2

These are the values taken at t = 0.

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Reflectionless potentials: Here we have R = 0 simply. We still have tonormalize the eigenfunctions though. They are

ψ = AP µN (312)

where µ = 1 . . . N. When µ is integer we find that∫|P µN |

2dx =(N + µ)!

µ (N − µ)!. (313)

In addition the asymptotics of P µN (which after some tedious effort we

can get from the series expansion of the hypergeometric function) is

P µN ∼

(N + µ)!

µ! (N − µ)!e−µx (314)

so that the normalized eigenfunction goes as

ψµN ∼1

µ!

√µ (N + µ)!

(N − µ)!e−µx (315)

and therefore

cµN =1

µ!

√µ (N + µ)!

(N − µ)!(316)

e.g. when N = 2 we get c12 =√

6 and c22 =√

12.

Almost reflectionless potentials: This corresponds to the u = 2.001sech 2(x)example, where we expect to get a very small oscillatory component (thedispersive wave that is initially shed by the soliton). We can adapt theprevious case using a small perturbation. Let

a = −N ′(N ′ + 1)

where N ′ = N + ε. The eigenvalues are µ′ = N ′ −M = µ + ε whereM is integer and since εis small we can set M = 1 . . . N − 1 again. Thenormalization of eigenfunction proceeds exactly as before, except that wemust replace factorials with Γfunctions. The expansion of a Γ−functiongoes like

Γ(N ′) = Γ(N)(1 + εψ(N) + . . .) (317)

where ψ = Γ′/Γ is called the digamma function (you don’t need to knowanything about it). Expanding the terms and retaining the first ε pieceI get

cµ′

N ′ = cµN(1 + ε[1

µ+ψ(N + µ+ 1)

2− ψ(µ+ 1) ]). (318)

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R(k) is easier and probably more interesting;

R(k) =Γ(−ik −N ′)Γ(1− ik +N ′)Γ(ik)

Γ(−N ′)Γ(1 +N ′)Γ(−ik)

=sin(πN ′)

π

Γ(−ik −N ′)Γ(1− ik +N ′)Γ(ik)

Γ(−ik)

≈ εΓ(−ik −N)Γ(1− ik +N)Γ(ik)

Γ(−ik). (319)

It will be our dispersive wave component eventually.

14.3 Evolving the data.

Now we have collected our data we need to evolve it forwards in time. Forthe moment I will outline the straightforward brute force method in Drazinand Johnson and concentrate manly on the discrete data. In the next sectionwe will see the smarter and much more profound Lax pair method, which alsoteaches us something about why this method works at all.

Begin with the time independent SE. I’ll reinstate the u = −V since we ’llneed at some point to use the KdV equation for this. The equation is

−ψxx + (λ− u)ψ = 0 (320)

and we want to show that λt = 0 if u evolves through time according to theKdV equation.

Proof: First differentiate the SE by x and t,

−ψxxx − uxψ + (λ− u)ψx = 0

−ψxxt + (λt − ut)ψ + (λ− u)ψt = 0.

I stress that ψ is here differentiated w.r.t. time even though it is thetime-independent SE that it solves - it can at the moment be a functionof time purely because of time appearing as a parameter in u, λ.

Now defineK(x, t) = ψt − uxψ + 2(u+ 2λ)ψx (321)

and construct the identity

∂x(ψxK − ψKx) = ψxx(ψt − uxψ + 2(u+ 2λ)ψx)

− ψ(ψxxt − uxxxψ + 3uxψxx + 2(u+ 2λ)ψxxx).(322)

(Note that the sign differences with D+J are due to their opposite signuux term in the KdV equation. This expression and the next are tedious

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to get - use mathematica or maple if you want to check them.) Nextsubstitute for ψxxxand ψxxt and ψxx gives

∂x(ψxK − ψKx) = ψ2(−λt + ux + 6uux + uxxx)

= −λtψ2 (323)

where in the last line I used the fact that u satisfies the KdV equation.Finally we note that the LHS is a total derivative. Thus we can integrateby x to get

[ψxK − ψKx]∞−∞ = −λt

∫ψ2dx. (324)

Consider the bound-state eigenfunctions that have λ = µ2 > 0 (i.e.negative energy) then the eigenfunctions are real and are normalizedsuch that

∫ψ2dx = 1. In addition they exponentially decay at infinity,

so that Lim|x|→∞(K,ψ, ψx, Kx) = 0 and hence λt = 0.

14.3.1 Evolution of the Discrete spectrum

Thus the eigenvalues are constants of the motion, and asymptotically the dis-crete eigenfunctions have to evolve as

ψ ∼ cn(t)e−µn|x| ; |x| → ∞ (325)

where now µn is constant. Having established this fact we can work with Kto get the time evolution of the scattering data. Integrating by x we have

ψxK −Kxψ = 0 (326)

where the RHS is zero (not a time dependent x-integration constant) becauseof the boundary condition that the LHS vanishes at spatial infinity. A furtherintegration gives

K = hλ(t)ψ (327)

where hλ is the t−dependent integration constant that can in principle bedifferent for every λ. However it turns out (see D+J for details – insert K anduse the SE and integrate again) that hλ = 0 is the only choice consistent withthe boundary conditions of ψ,K vanishing at spatial infinity. Consequently

K ∼ 0 = ψt − uxψ + 2(u+ 2λ)ψx (328)

This is our evolution equation for ψ and hence our scattering data.We can now insert the asymptotic form of ψ of eq.325 (along with the

asymptotic boundary condition u, ux → 0) into the above to find

(cn)t − 4µ3ncn = 0

cn(t) = cn(0)e4µ3nt.

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14.3.2 The continuous spectrum

For λ = −k2 < 0, we have a continuous spectrum. We may follow the timeevolution of the reflection and transmission coefficients of the continuous eigen-functions by keeping λ = −k2 fixed. First we find the asympotic form of Kfrom its definition in eq.321

K ∼ (Tt − 4ik3T )eikx ; x→ +∞K ∼ (Rt + 4ik3R)e−ikx − 4ik3eikx ; x→ −∞ (329)

Now inserting this into eq.324 and insisting that λt = 0 gives us

Rt + 8ik3R = 0. (330)

To get T we need to integrate eq.324.

ψxK − ψKx = gk(t) (331)

but since this is true for all x and we have just chosen R so the RHS vanishesat the x→ −∞ limit we must have gk = 0. Integrating again

K = hk(t)ψ (332)

In the x→ −∞ limit

hk (Re−ikx + eikx) = (Rt + 4ik3R)e−ikx − 4ik3eikx

= −4ik3(Re−ikx + eikx)

determining hk = −4ik3. In the x→ +∞ limit we find

Tt − 4ik3T = hk T

→ Tt = 0. (333)

Taken together then we can summarize the evolution of the scattering data as

µn = constant

cn(t) = cn(0)e4µ3nt

R(k, t) = R(k, 0)e−8ik3t

T (k, t) = T (k, 0) (334)

Note that the change in R is a phase so that we still have |T |2 + |R|2 = 1.

89

15 (Long) Digression: Lax pairs and the KdV

hierarchy

The analysis in the previous section can be improved because

• It was rather clumsy and the importance of the KdV equation was a bitopaque

• We would like to be able to see if there are other equations that have thesame isospectral properties and can also be solved using inverse scatter-ing

These questions can be addressed using the Lax pair formalism. The result isthe discovery of an infinite hierarchy of equation that can be solved by inversescattering like the KdV equation, and a deep connection with the conservedcurrents discussed earlier.

15.1 The idea of a Lax pair

Consider the eigenvalue problem of the time independent SE

(∂2x + u)ψ = λψ (335)

with ψ remaining bounded ∀x. The Lax method concentrates on the propertiesof the operator L = ∂2

x + u so that

Lψ = λψ. (336)

I’m going to generalize the discussion so that later we can derive equationssimilar to the KdV equation, so I’ll say that the equation we want to solve is

ut = N(u) (337)

where for the KdV eqn we have N(u) = −(6uux + uxxx) but in principle Ncan be any function of u. What we would eventually like to know is what formof N gives isospectral evolution of eigenfunctions. The key properties are

1. The set of eigenvalues λ of L(u) is independent of t

2. There is a set of eigenfunctions ψ of L that evolves in t as ψ(t) = Mψ,where M is another time-independent differential operator

We have seen property (1) for the KdV equation already, but we still need toconfirm property (2) as we have so far been concentrating on the asymptoticbehaviour.

First assume that ∃ an operator M such that the equation ut = N(u) canbe written

Lt + [L,M ] = 0 (338)

90

Here all of these objects are operators, but the [L,M ] term should be “mul-tiplicative” if the equation is to make sense - this will become clear shortly.Also the t differentiation is on the u dependence in L. One has to imagine allof the operators acting of functions of x so really it should be

Ltψ(x) + [L,M ]ψ(x) = 0. (339)

1Now let’s use the eigenvalue equation to determine λt by differentiating (L−λ)ψ

λtψ = Ltψ + Lψt − λψt

= (ML− LM)ψ + (L− λ)ψt

= (Mλ− LM)ψ + (L− λ)ψt

= (L− λ)(ψt −Mψ) (340)

Now multiply by ψ∗ and integrate over x gives,

λt

∫ ∞

−∞|ψ|2dx = λt =

∫ ∞

−∞ψ∗(L− λ)(ψt −Mψ) dx (341)

where I assumed a conventionally normalized eigenfunction∫∞−∞ |ψ|

2 dx = 1.At this point I’ll introduce an important property of L, self-adjointness.

Definition: An operator L is called self-adjoint if for any pair of functionsη(x), χ(x) it obeys ∫ ∞

−∞η∗Lχdx =

∫ ∞

−∞(Lη)∗χdx (342)

Proof: to prove L is self adjoint integrate by parts twice∫η∗(∂2

x + u)χdx = surface− terms +

∫(∂2xη

∗ + uη∗)χdx

=

∫ ∞

−∞(Lη)∗χdx

where all the surface terms vanished in the x→ ±∞ limit.

1For example if L = ∂x and M = u+ ∂x (which they don’t but never mind) then

[L,M ]ψ = (LM −ML)ψ= ∂x(uψ + ∂xψ)− (u∂xψ + ∂2

xψ)= uxψ.

This is multiplicative in the sense that there is no ∂xψ term. This is all longhand. Inshorthand it is easier to drop the ψ and write

[L,M ] = (LM −ML)= ∂xu+ ∂2

x − u∂x − ∂2x

= ux + u∂x − u∂x = ux

where the last line follows once you remember to let the differential operators act on allfunctions to the right, i.e. u and the unwritten ψ.

91

Using this result, eq.341 becomes

λt =

∫ ∞

−∞((L− λ)ψ)∗(ψt −Mψ) dx

= 0 (343)

where we used (L − λ)ψ = 0 in the last line. This completes the proof thatλt = 0 if the operator L obeys Lt + [L,M ] = 0.2 We have yet to determine thetime evolution of ψ. Having established that λt = 0 we can return to eq.340

(L− λ)(ψt −Mψ) = 0 (346)

This is in the form of an eigenvalue equation, and (because for each λ theeigenfunctions are unique) we must have

(ψt −Mψ) = aψ (347)

where a is some constant. However we can define a new M given by

M = M + a

for which the above givesψt = Mψ (348)

But in terms of M we have

Lt + [L, M ] = 0 (349)

so we may as well have used M instead of M in the first place.

2Generalization: The above discussion can be generalized. For example the operatorsL and M could be matrix valued. The generalization is then that the operators L,M acton some Hilbert space, and we work with the inner product (η, χ). For the example abovewe would define (η, χ) =

∫η∗χdx and the Hilbert space is the space spanned by normalized

eigenfunctions, but in general the inner product may involve a trace of matrices, or whatever.(You may have seen the equivalent in QM as 〈η|χ〉). In this notation self adjointess is written

(η, Lχ) = (Lη, χ) (344)

and the argument would have been written

λt(ψ,ψ) = (ψ, (L− λ)(ψt −Mψ))= ((L− λ)ψ, (ψt −Mψ))= 0. (345)

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15.2 The Lax pair for KdV

All that needs to be done now is to find correct M for the differential equation,in this case the KdV equation. It is possible to be systematic, but at this stageI’ll just present it and confirm it works; the Lax pair is

L = ∂2x + u

M = −(4∂3x + 6u∂x + 3ux). (350)

In order to avoid injury, it is conventional to define D = ∂x. Remember thateverything has the sense of operators acting on functions to the right so thatfor example

[u,D]f = (u∂x − ∂xu)f

= u∂xf − ∂x(uf)

= u∂xf − (∂xu)f − u(∂xf)

= −(∂xu)f (351)

so we would just write [u,D] = −ux. Now for the calculation. First, obviously

Lt = ut.

Then we have [L, M ] = [D2+u,−(4D3+6uD+3ux)]. Note that [Dn, Dm] = 0.Also we can expand the commutators to make things a bit easier before westart, and then note that the commutations involving [Dn, f(x)]acts essentiallylike a straight differentiation where you drop the fDn term and keep trailingDn−m terms; so

[L,M ] = 4[D3, u] + 6[uD, u]− 6[D2, u]D − 3[D2, ux]

= 4(uxxx + 3uxxD + 3uxD2)

+6uux − 6(uxx + 2uxD)D

−3(uxxx + 2uxxD)

= uxxx + 6uux.

Note that a crucial property is that the end result is “multiplicative”, all theterms with trailing D and D2 cancelled. That is [L,M ]when acting on somefunction f(x) just multiplies it by (6uux + uxxx). This will be very importantwhen it comes to making a more systemtic search for similar KdV-like systemslater. So finally

Lt + [L,M ] = ut + 6uux + uxxx

as required.

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15.3 Time evolution of scattering data revisited

Let’s check that the Lax pair gives the correct time evolution of the scatteringdata. The equation is

ψt = Mψ. (352)

For the scattering data we need only get M in the asymptotic (|x| → ∞)region;

M = −(4∂3x + 6u∂x + 3ux)

∼ −4∂3x (353)

where we used that u, ux → 0 at spatial infinity. For the reflection and trans-mission coefficients, we have

ψ ∼{A−(t)eikx +B−(t)e−ikx x→ −∞A+(t)eikx x→ +∞ (354)

I have set B+ = 0 (i.e. there is never any left moving wave on the right sideof the scattering region) but the normalization A− I leave as time dependent.Inserting this into eq.352 we get

∂tA−eikx + ∂tB−e

−ikx = 4ik3A−eikx − 4ik3B−e

−ikx x→ −∞∂tA+e

ikx = 4ik3A+eikx x→ +∞ (355)

Equating coefficients of e±ikx we get

A− = A−(0)e4ik3t

A+ = A+(0)e4ik3t

B− = B−(0)e−4ik3t. (356)

Finally to get R(t, k) and T (t, k) we need to divide the solution by A−(t) tonormalize the incoming wave to one. This gives R(t, k) = R(0, k)e−8ik3t andT (t, k) = T (0, k) as we found earlier.

For the time development of cn, we have asymptotically

ψ ∼ cn(t)e−µn|x|. (357)

Inserting this into eq.352 we get

∂tcn = 4µ3ncn (358)

so as found earlier,cn(t) = cn(0)e4µ

3nt (359)

Note that for the scattering data we only need the asymptotic formof ψ and the u, ux dependence dropped out of M . If this hadn’thappened the method would not have worked (i.e. we would havehad to have known the form of u(t) already in order to make theinverse transform).

94

15.4 The KdV hierarchy

It is natural to ask now if the KdV equation was the only equation withisospectral properties. The answer is no, and in fact the Lax pair methodshows us how to construct and infinite series of such equations.

Key point: the proof of an invariant spectrum only used the property thatthe evolution of L(u) can be written as

Lt + [L,M ] = 0 (360)

but didn’t actually rely on the form of M . In fact we required M =−(4D3 + 6uD + 3ux) to get the KdV equation, but this wasn’t used inthe proof.

The idea is to pick different M(u) such that the L evolution can be writtenin the same way but the equation generated doesn’t have to be the KdVequation. This will be enough to guarantee the constancy of the spectrum(i.e. time independence of the eigenvalues, λt = 0 ).

However M(u) is not arbitrary. Since Lt = ut is purely multiplicative thenso must [L,M ] be. That is all the D’s must cancel in [L,M ].If they do thenLt + [L,M ] is the required equation for u as a p.d.e. Let’s try some examples.In what follows I’ll always be assuming L = D2 + u;

Example 1: M = α(x) for some function α. Then

[L,M ] = [D2 + u, α]

= [D2, α]

= αxx + 2αxD (361)

So in this case we require αx = 0 or α = constant. But then αxx = 0 aswell and Lt = ut = 0 is the final equation. So this certainly has λt = 0but it is a bit trivial.

Example 2: M = α(x)D + β(x) for some functions α and β. Then

[L,M ] = [D2 + u, αD +B]

= [D2, B] + [D2, α]D − α[D, u]

= βxx + 2βxD + αxxD + 2αxD2 − αux.

So in this case we need 2βx +αxx = 0 and αx = 0. Again we have α andβ both constant so αxx = βxx = 0, but

Lt + [L,M ] = ut − αux = 0. (362)

This is a linear equation that can again be trivially solved by D’Alembert’ssolution. i.e. any function f(x + αt) solves it. Again it is obvious thatthe eigenvalues remain constant because the initial “wave” profile of uis just translated. This is maybe the first hint that the simplicity of theKdV equation and the linear equation is in some sense related.

95

As an exercise we can check that the eigenfunctions evolve according to

ψt = M(u)ψ

= αψx + βψ

Let z± = t± αx. Then this can be rewritten as

ψz−ψ

= β

orψ = exp(βz− + γ(x+))

where γ is any function of t+αx that has to be set by the initial scatteringdata.

Example 3: The final example to try “by hand” is M = α(x)D3 + β(x)D +γ(x) with α, β, γ again being arbitrary functions of x.Then

[L,M ] = [D2 + u, αD3 + βD + γ]

= [D2, α]D2 + [D2, β]D + [D2, γ]

−α[D3, u]− β[D, u]

= αxxD2 + 2αxD

3 + βxxD + 2βxD2 + γxx + 2γxD

−α(uxxx + 3uxxD + 3uxD2)− βux (363)

Setting the coefficients of D3, D2 and D equal to zero we get

2αx = 0

αxx + 2βx − 3αux = 0

βxx + 2γx − 3αuxx = 0 (364)

The first relation gives α = constant, so αx = αxx = 0. The second thengives

β =3

2αu+ c

where c is a constant. The third gives

γ =3

4αux + d

where d is another constant. Substituting into eq.363 we find

Lt + [L,M ] = ut −α

4uxxx −

3α

2uux − cux.

Choosing α = −4 and c = 0 gives the KdV equation! For more generalvalues we can just rescale the coordinates; you can check that

u(x, t) = u(x+αc

4t, −α

4t)

satisfies the KdV equation. This shows for the first time that the KdVequation is in some sense “natural”, and is one of the lower members ofa hierarchy” of such equations.

96

15.4.1 Hints for the general case

In principle we could keep going, but the calculations are getting pretty grim.Lets look at some hints about how to simplify things a little. We need somenew technology.

1. Hermitian inner product: (η, χ). For the example above we woulddefine (η, χ) =

∫η∗χdx and the Hilbert space is the space spanned by

normalized eigenfunctions, but in general the inner product may involvea trace of matrices, or whatever. (You may have seen the equivalent inQM as 〈η|χ〉). The argument for λt = 0 would have been written

λt(ψ, ψ) = (ψ, (L− λ)(ψt −Mψ))

= ((L− λ)ψ, (ψt −Mψ))

= 0. (365)

Now we take into account that the ψ and η may be matrix valued orvectors.

2. The adjoint of an operator: Before we used the property of self-adjointness of L. We can define the “adjoint” of L with a dagger suchthat as

(η, Lχ) = (L†η, χ) (366)

Then self-adjointness is the condition that

L† = L (367)

so that(η, Lχ) = (Lη, χ) (368)

as required. The dagger is the like the hermitian conjugate of a matrix.The analogy is with a matrix M and vectors U, V where you can checkthings work in the way you’d expect. The hermitian inner product wouldbe (V, U) = V †.U where now (V † = (V ∗)T ). You can check that

(M †V, U) = (M †V )†U

= V †MU

= (V, MU) (369)

and self-adjointness means that the matrix is hermitian M = M †. Withrather sloppy nomenclature self-adjoint operators are often called sym-metric, L† = L. The opposite is called anti-symmetric or skew-symmetricM † = −M .

Now note the condition that [L,M ] is a real, multiplicative function and con-tains no operators. By definition of self-adjointness we must have

[L,M ] = [L,M ]†. (370)

97

What can we learn about M from this? Using the rules in appendix D we cando the following

[L,M ] = (LM)† − (ML)†

LM −ML = M †L† − L†M †

LM −ML = M †L− LM †

−→ [L, M +M †] = 0 (371)

Just as for matrices we can decompose M into symmetric and antisymmetricparts as follows

MS =(M +M †)

2

MA =(M −M †)

2.

The above equation tells us that the symmetric part of M commutes with L.Since the equation of interest is Lt + [L,M ] = 0 we may as well take M to beantisymmetric.

15.4.2 How to write M?

Instead of writing the general M of order m as∑m

j αj(x)Dj where αj are some

functions of u ux etc, we can choose a different basis by choosing

m∑j

βjDj +Djβj (372)

where βj are some other functions. Now by integration by parts we have

D† = −D and of course the functions obey β†j = βj. So then

(D2k)† = D2k

(D2k−1)† = −D2k−1 (373)

so we may take the odd powers of D only

m∑k=1

β2k−1D2k−1 +D2k−1β2k−1 (374)

Consider the KdV operator

M(u) = −4D3 − 6uD − 3ux

= (−2D3 − 2D3) + (−3uD − 3Du ) (375)

so the above is a consistent way to write is. There is one additional simplifyingfeature; if [L,M ] is multiplicative, then the leading term has β2m−1 = constant.(as an exercise show this.)

98

Now there is no alternative but hard work. The constraint of multiplicativeoperators means that the vanishing of each term in Dmin M imposes exactlyone constraint, so that there are exactly as many constraints as functions βjand therefore the latter are completely constrained. Let’s look at the first fewexamples;

1. m = 1; the highest power is 0 so trivially

ut = 0 (376)

.

2. m = 2; the highest power is 1, we have M = −D, (i.e. recall that theleading β is a constant) and

[L,M ] = −[D2 + u, D]

= ux

So ut + ux = 0

3. m = 2; the highest power is 3 and we have M = −D3 + β1D +Dβ1.

[L,M ] = [D2 + u, −D3 + β1D +Dβ1]

= [D2, β1]D +D[D2, β1] + [D3, u]− β1[D, u]− [D, u]β1

= β1xxD + 2β1xD2 +D(β1xx + 2β1xD) + uxxx + 3uxxD + 3uxD

2 − 2β1ux

= β1xxD + 2β1xD2 + (β1xx + 2β1xD)D +

(β1xxx + 2β1xxD)− uxxx − 3uxxD − 3uxD2 − 2β1ux

= (4β1xx − 3uxx)D + (4β1x − 3ux)D2 + β1xxx − uxxx − 2β1ux (377)

So that D and D2 vanishing implies β1 = 34u. Inserting this we get

0 = 4ut − uxxx − 6uux.

A simple rescaling (e.g. t → −4t or M → −4M) yields the usual KdVequation.

4. m = 3 : The highest power is 5. This gets messy (left as characterbuilding exercise)

ut + 30u2ux + 20uxuxx + 10uuxxx + uxxxxx = 0 (378)

This is a new 5th order equation with the same properties as KdV. The seriescontinues to infinity.

99

16 Connection with conservation laws

There is an amazing connection with the conserved charges and conservationlaws. Recall the the KdV equation has an infinite sequence of charges, givenby

Qn =

∫ ∞

−∞Tndx

where (Tn)t + (Xn)x = 0 for some Xn with [X]∞−∞ = 0. The series is

T1 = u

T2 = u2

T3 = u3 − 1

2u2x

T4 = u4 − 2uu2x +

1

5u2xx

T5 = u5 − 105

21u2u2

x + uu2xx −

1

21u2xxx

Tn = un + . . . (379)

where T4,5 are new to you. We now have two infinite sequences

• For the KdV eqn itself an infinite sequence of T ′ns

• Going beyond the KdV eqn, an infinite sequence of Nn(u) such thatut = Nn(u) leaves the eigenvalues of L = ∂2

x + u invariant

How do these two sequences tie together if at all? The most boring possibilityis that each Nn(u) has its own completely separate sequence of T ′ns howeverthe answer is much more mysterious.

16.1 Functional or Frechet derivatives

To connect the two sequences we need an extra tool, the functional derivative.Recall from the variational idea from Lagrangian Mechanics (see AppendixB). We first defined the Lagrange density , L such that

S[u] =

∫ t2

t1

∫ ∞

∞L(u, ut, ux, uxx, uxxx, . . .) dxdt (380)

I have here generalized things to allow dependence on higher derivatives of uthan you would normally consider. Let u → u + δu then if S → S + δS wehave

δS = 0 =

∫ ∫L(u+ δu, ut + δut, ux + δux, . . .)− L(u, ut, ux) dtdx

=

∫ ∫∂L∂u

δu+∂L∂ut

δut +∂L∂ux

δux +∂L∂uxx

δuxx +∂L∂uxxx

δuxxx + . . . dtdx

=

∫ ∫ (∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

+d2

dx2

∂L∂ux

− d3

dx3

∂L∂ux

+ . . .

)δu dtdx.(381)

100

Here we can got the last line by repeatedly integrating by parts in the relevantintegrand, x or t, and there is a plus or minus sign depending on whetherwe need an even or odd number of integration by parts. The main point wasto get everything multiplying δu which leads to the Euler-Lagrange equationsupon setting the function inside the brackets to zero. The object inside thebrackets is defined to be the functional derivative of S with respect to u.

δS

δu=∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

+d2

dx2

∂L∂ux

− d3

dx3

∂L∂ux

+ . . . (382)

so that the concept of derivatives (in particular f(x + δx) = f(x) + dfdxδx) is

generalized to integrals of functions;

S[u+ δu] = S[u] +

∫δS

δuδu dxdt+O(δu2). (383)

We can also find functional derivatives for integrals just over x. e.g. if

F [u] =

∫f(u, ux, uxx . . .) dx (384)

then

F [u+ δu] = F [u] +

∫ (∂f

∂u− d

dx

∂f

∂ux+

d2

dx2

∂f

∂ux− d3

dx3

∂f

∂ux+ . . .

)δu dx.

= F [u] +

∫ (δF

δu

)δu dx + O(δu2) (385)

The only difference to the action principle is the disappearance of the “time”-pieces, so actually this is simpler than the action principle with which you maybe more familiar. Some examples:

f = u ⇒ δFδu

= ∂f∂u

= 1

f = u3 ⇒ δFδu

= ∂f∂u

= 3u2

f = u2x ⇒ δF

δu= ∂f

∂u− d

dx∂f∂ux

= −2uxx

One additional feature that will become important in a minute is that if f isa total derivative (i.e. f = ∂xg) then

δF

δu= 0. (386)

The proof of this I’ll leave as an exercise, but it is clearly true for examplefor g being a function of u only; by the chain rule

f = uxguδF

δu= uxguu − ∂xgu

= uxguu − uxguu

= 0

Now consider the conserved quantities, Qn =∫Tndx

101

δQ1

δu= δ(u)

δu= 1

δQ2

δu= δ(u2)

δu= 2u

δQ3

δu=

δ(u3− 12u2

x)

δu= 3u2 + uxx

δQ4

δu=

δ(u4−2uu2x+ 1

5u2

xx)

δu= 4u3 + 4u2

x + 4uuxx − 2u2x + 2

5uxxxx

δQ5

δu=

δ(u5− 10521u2u2

x+uu2xx− 1

21u2

xxx)

δu= uurgh

Taking the derivatives with respect to x gives

∂x(δQn

δu) = 0, 2ux, 6uux + uxxx,

1

5(uxxxxx + 10uuxxx + 20uxuxx + 30u2ux), . . .

(387)This should be compared with the Nn(u) operators (rescaled)

Nn(u) = 0, ux, , 6uux + uxxx , uxxxxx + 10uuxxx + 20uxuxx + 30u2ux

So the two sequences are exactly the same in this sense. There are someother remarkable facts

• If u evolves by the mth KdV equation exactly the same set of Q′ns are

conserved quantities

• Imagine we have one time for each equation t1, t2 . . . etc, so that we haveu(x, t1, t2 . . .), and that evolution through tm is given by Nm, so we havea set of evolution equations;

utm = Nm(u).

Then if we evolve for a while in tn and then in tm6=n, it doesn’t matterwhich way round we do it. This is the idea of commuting flows, and is avery important aspect of “modern” soliton theory.

102

17 The inverse scattering transform

Finally we have to reconstruct the potential at time t by performing the inversescattering transform. We have collected the data

µn(t) = µn(0)

cn(t) = cn(0)e4µ3nt

R(k, t) = R(k, 0)e−8ik3t (388)

confident that we can determine the form of u(t) from it, but is this possible?In other words, if you are allowed to sit at spatial infinity and bounce particlesof the potential, can you tell the shape of the potential by just measuring thedata. This is a question of some practical importance, for example in sonar.

For the 1d SE the answer was already well known going back to Gel’fandand Levitan in 1951 and Marchenko in 1955. The recipe for reconstructingu(t) is as follows (I’ll use it without proof - for the proof see D+J).

17.1 The Marchenko equation

To reconstruct u(x, t) from the scattering data

1. construct the function

F (ξ) =N∑n=1

c2ne−µnξ +

1

2π

∫ ∞

−∞R(k)eikξdk (389)

where N is the number of solitons in the spectrum, and both cn, R(k)are to be evaluated at time t.

2. Solve the following linear integral equation for K(x, y)

K(x, z) + F (x+ z) +

∫ ∞

x

K(x, y)F (y + z) dy = 0 (390)

3. u(x, t) (which is equal to−V (x, t)) is given by

u(x, t) = 2∂xK(x, x). (391)

The equation may or may not be solvable, but the main point is that theinverse problem is essentially the same at all times, with the time dependencein F being given by

F (ξ; t) =N∑n=1

c2n(0)e8µ3

nt−µnξ +1

2π

∫ ∞

−∞R(k, 0)eikξ−8ik3tdk (392)

Taken together the technique is a remarkable discovery, and is a kind of “non-linear Fourier analysis”. To make the final step clearer we’ll look at someexamples.

103

17.2 The single KdV soliton

Returning to the single soliton case; do we recover the travelling wave solution?Here as discussed our starting function is u(0) = 2sech 2x. Inserting into thecµN equation 316 with N = µ = 1 we have simply

c11 =√

2 (393)

and R(t, k) = 0, so that

F (ξ, t) = 2e4t−ξ (394)

The Marchenko equation is

K(x, z) + 2e8t−(x+z) + 2

∫ ∞

x

K(x, y) e8t−(y+z)dy = 0 (395)

Since y is just a dummy variable, we can extract the e−z factors and theequation is in the form

K(x, z) + e−z (2e8t−x + 2

∫ ∞

x

K(x, y) e8t−ydy) = 0 (396)

The solution must clearly be in the form K(x, z) = J(x)e−z where

J(x) = −2e8t−x − 2J(x)

∫ ∞

x

e8t−2ydy

= −2e8t−x − J(x)e8t−2x

so that

K(x, z) = − 2e8t−(x+z)

1 + e8t−2x. (397)

So that

u(x, t) = 2∂xK(x, x)

= −4∂x(e2x−8t − 1)−1

= 8e2x−8t(e2x−8t − 1)−2

= 2sech 2(x− 4t) (398)

which is the travelling wave solution!

17.3 The almost reflectionless case; u(0) = 2.001sech 2x

In this case it is possible to develop a perturbation series to find solutions. Pre-viously we found that the scattering data for the almost reflectionless potential(that is u = N ′(N ′ + 1)sech 2x where N ′ = N + ε) can be written

cµ′

N ′(0) = cµN(1 + εα(µ,N))

R(k, 0) = εβ(k,N)

104

where α, β are some not very nice functions of N,µ, k. We can do a perturba-tion series in ε by defining K = K0 + εK1 + . . . and F = F0 + εF1 + . . . . TheMarchenko equation can then be solved iteratively; i.e. the full equation plusthe zeroth order solution we have just found

K0(x, z) + F0(x+ z) +

∫ ∞

x

K0(x, y)F0(y + z) dy = 0

=⇒ K1(x, z) + F1(x+ z) +

∫ ∞

x

(K1(x, y)F0(y + z) +K0(x, y)F1(y + z)) dy = 0(399)

Since this equation is linear in expontentials things become more tractablealthough not very pleasant (especially the second term in the integral). Havea go at the N = µ = 1 single soliton case using

α(1, 1) =1

4(3 + 2γE)

β(k, 1) = −Γ(2− ik)

1 + ik

where γE = 0.577216. If you can solve it you should find a dispersive com-ponent travelling to the left. It probably helps to swop the order of k and yintegrations, but I suspect you’ll still be left with a single difficult integral todo.

17.4 The N-soliton solution

The single soliton solution indicates how to proceed with the multi-solitonone, and we actually get new results. Begin by constructing F again; it canbe written as

F (ξ; t) =N∑µ=1

(cµN(0))2e8µ3t−µξ (400)

since as we have seen R = 0 ∀t. The Marchenko equation is in the form

K(x, z) +∑

(cµN(t))2e−µ(x+z) +∑µ

(cµN(t))2

∫ ∞

x

K(x, y) e−µ(y+z)dy = 0 (401)

We now define a N component vector consisting of entries e−µz with µ =1 . . . N , which I’ll call η. Now suppose the solution is in the form

K(x, z) =∑µ

J(x)µηµ(z) (402)

where J is now another N vector. The Marchenko equation is now∑µ

Jµηµ(z) =∑µ

(cµN(t))2ηµ(z) (e−µx +∑ν

Jν

∫ ∞

x

e−(µ+ν)ydy)

=∑µ

(cµN(t))2 ηµ(z) (e−µx +∑ν

Jνe−(µ+ν)x

µ+ ν) (403)

105

Following our previous success, look for a solution where all the coefficientsvanish for every ηµ. The above equation becomes a matrix equation

Jµ = (cµN(t))2 (e−µx +∑ν

Jνe−(µ+ν)x

µ+ ν) (404)

which can be writtenJ = S−1

N L (405)

where

SNµν = δµν − (cµN(t))2 e−(µ+ν)x

µ+ ν

Lµ = (cµN(t))2 e−µx (406)

To find the solution we now note that

∂xSNµν = (cµN(t))2 e−(µ+ν)x

= Lµe−νx

soJµ =

∑ν

(S−1N )µν(∂xSN)νµe

µx (407)

and

K(x, x) =∑µ

Jµe−µx

=∑µν

(S−1N )µν(∂xSN)νµ

= Tr(S−1N ∂xSN) (408)

This can (may) be recognized formally as

∂xTr(log(SN)) = ∂x log(detSN).

Finally the u(x, t) can be written as

u(x, t) = 2∂xK(x, x)

= 2∂2x log(detSN) (409)

As an exercise you can show that this is equivalent to the previous N-solitonsolution we wrote derived using Hirota’s method.

106

18 Integrable systems in classical mechanics

The systems we have been looking at so far can be thought of as an integrableinfinite Hamiltonian system - it’s a one dimensional field theory which can bethought of as an infinite number of coupled oscillators (c.f. the Sine-Gordondiscussion). Not surprisingly then, Lax pairs can be applied to finite systemsas well, and there are various famous examples of completely integrable systemsin classical mechanics. A Hamiltonian system is define by a set of coordinatesqk=1...n momenta pk=1...n and a Hamiltonian, H(q, p) defined such that

q =∂H

∂p

p = −∂H∂q

(410)

Definition: A hamiltonian system H(qk, pk) where k = 1 . . . n is called com-pletely integrable if it has n integrals of motion Qk(q, p) where {H,Qk} =0 which are time independent, and in mutual involution {Qk, Qj 6=k} = 0.

The interest for our discussion is that the integrability of classical systems canbe shown by the existence of a Lax pair, which allow it to be solved. And theintegrals of motion are precisely the eigenvalues of the Lax equation;

L = [L,M ] (411)

(I’ll use a dot notation for time derivatives, L ≡ Lt) To clarify what the formof L or M should be consider some examples.

18.1 The Lax pair for the simple harmonic oscillator

In the field theory, we saw that there Lax pair equation involved differentialoperators acting on function. In finite systems this is replaced by matrixequations instead (e.g. think of the differential operator before taking thelimit, it would be (qk+1 − qk)/ε, and think of qk as being an n−vector). Thesimplest example is the single simple harmonic oscillator with Lax pair

L =

(p ωqωq −p

)M =

(0 −ω

2ω2

0

)(412)

The Lax equation becomes(p ωqωq −p

)=

(−ω2q ωpωp ω2q

)(413)

107

From this we can read off q and p and find

q = p

p = −ω2q (414)

which gives H(q, p) = 12p2 + ω2

2q2. There is a formal solution to the Lax

equation given byL(t) = U(t)L(0)U(t)−1 (415)

where U is a solution to the initial value problem

U = −MU

U(0) = 1 (416)

I’ll assume without proof that the inverse of U exists. It is easy to differentiatethe L(t) solution to see that L(t) indeed satisfies the Lax equation;

L = UL(0)U−1 − ULU−1UU−1

= −MUL(0)U−1 + UL(0)U−1M

= −ML(t) + L(t)M (417)

The formal solution to the IVP for U is given by

U(t) = e−Mt. (418)

All we need to know is what exponential of a matrix means! The easiest wayto the answer is to Taylor expand the exponential

eMt = 1 +Mt+M2t2

2!+ . . . (419)

It is easy to check that

Mn = (ω

2)n

(−1)

n−12

(0 −11 0

)n− odd

(−1)n2

(1 00 1

)n− even

(420)

so resumming the series in each element we have

eMt =

(cos ωt

2− sin ωt

2

sin ωt2

cos ωt2

)(421)

giving

L =

(cos ωt

2− sin ωt

2

sin ωt2

cos ωt2

) (p0 ωq0ωq0 −p0

) (cos ωt

2sin ωt

2

− sin ωt2

cos ωt2

)(

p(t) ωq(t)ωq(t) −p(t)

)=

(p0 cosωt− ωq0 sinωt p0 sinωt+ ωq0 cosωtp0 sinωt+ ωq0 cosωt −p0 cosωt+ ωq0 sinωt

)108

from which we can read off

q(t) =p0

ωsinωt+ q0 cosωt (422)

the standard solution for the SHO. The fact that there is a Lax equation tellsus there must be a single constant of the motion which in this case is theHamiltonian itself; it can be written

H =1

4Tr(L2). (423)

It is easy to see that this should be a constant because by this definition

H =1

2Tr(LL)

=1

2Tr(L(LM −ML)) = 0 (424)

by the cyclic properties of the trace.

18.2 The Toda Lattice

The SHO is a rather trivial example ; in particular we can solve it extremelyeasily using usual techniques anyway. A less trivial example of a matrix systemwith a Lax pair solution is the Toda lattice, which describes the dynamics ofn− ”particles” (labelled k = 1 . . . n) on a line, interacting with the closest setof neighbours. The Hamiltonian is given by

H(p, q) =1

2

n∑k=1

p2k +

n−1∑k=1

g2ke

2(qk−qK+1) (425)

The system can be thought of as similar to the suspended oscillators on aline we used for the Sine-Gordon example, but with an exponential potentialwhose coupling gk depends on the position, k, so in this respect it is actuallyquite general.

Hamiltons equations (the equations of motion) are quite simple

qk =∂H

∂pk= pk

pk = −∂H∂qk

= 2(g2k−1e

2(qk−1−qk) − g2ke

2(qk−qk+1)) (426)

A bit of matrix manipulation shows that these equations are realized by Laxequation with the following Lax pair;

L =

p1 g1e

2(q1−q2)

g1 p2 g2e2(q2−q3)

g2. . . . . .. . . gn−1e

2(qn−1−qn)

gn−1 pn

109

M = −2

0 g1e

2(q1−q2)

0 0 g2e2(q2−q3)

0. . . . . .. . . gn−1e

2(qn−1−qn)

0 0

(427)

The conserved quantities follow as for the SHO by noting that

Qk =1

kTr(Lk) (428)

is independent of time since by the Lax equation

d

dtQk = Tr(Lk−1L) = Tr(Lk−1(LM −ML)) = 0 (429)

The solution is by no means as obvious as the SHO one; it was provided byOlshanetsky and Perelomov in 1979.

We wish to find the solution (i.e. the time dependent functions of qk(t)) toLt = [L,M ] with L,M defined as above. Begin by noting that every Hermitianmatrix (Y = Y †) has a unique decomposition

Y = ZXZ† (430)

where X is the diagonal matrix of real eigenvalues. We will eventually have

X =

e2q1

e2q2

. . .

e2qn

. (431)

and will take Y (0) = X(0) is the matrix of initial values of X, and Z encodeshow they mix up on time evolution with Z(0) = 1. Differentiating,

Y = ZXZ† + ZXZ† + ZXZ†

and

Y Y −1 = (ZXZ† + ZXZ† + ZXZ†)(ZXZ†)−1

= (ZXZ† + ZXZ† + ZXZ†)(Z†)−1X−1Z−1

= ZZ−1 + ZXX−1Z−1 + ZXZ†(Z†)−1X−1Z−1

= Z {Z−1Z + XX−1 +XZ†(Z†)−1X−1}Z−1 (432)

Now suppose that Z is the solution to Z = −ZM (and also by taking thehermitian conjugate Z† = −M †Z†) and substitute this into the above equation

Y Y −1 = Z {−M + XX−1 −XM †X−1}Z−1. (433)

110

It is straightforward to verify that

Y Y −1 = 2ZLZ−1. (434)

Now if we can determine Y as well then we should be able to get X which willgive us qk(t). Differentiating the above by t we find

d

dt(Y Y −1) = 2(ZLZ−1 + ZLZ−1 + ZLZ−1)

= 2Z {Z−1ZL+ L+ LZ−1Z}Z−1

= 2Z {−ML+ L+ LM}Z−1 = 0 (435)

So then we must haveY Y −1 = D (436)

where D is a constant n× n matrix, and

Y = DY

for some D. This equation can be integrated as usual (normally we wouldhave log(y/b) = d where b is some constant, but now we need simply note thatY is hermitian by definition) to give

Y = BeDtB† (437)

where Y (0) = BB†. The definition Y (0) = X(0)means that the B constant ofintegration is

B = X(0)12 (438)

i.e. the matrix of eqk(0) on the diagonals. Also Y (0)Y (0)−1 = 2L(0) gives theD constant of integration as

D = B−1L(0)B (439)

This determines y(t). So finally

Y (t) = ZXZ† = BeDtB†

=⇒ X(t) = Z−1BeDtB†(Z†)−1. (440)

111

Appendix A: Numerically solving p.d.e.s

The mathematica program that produced all the solution plots is as shownbelow. This can easily be modified to solve Sine Gordon, the dispersive equa-tion etc. There are a number of example starting waves, only one of whichcorresponds to the actual KdV soliton.

Clear@"�"Dkdv@8a_<, t_, x_D := D@p@t, xD, tD +

a p@t, xD D@p@t, xD, xD + D@p@t, xD, 8x, 3<D � 0;

cell = 40;s@8a_, c0_, x0_<, t_, x_D := 3 c0�a HSech@ Sqrt@c0D�2 Hx - x0 - c0 tLDL^2;s@8a_, c0_, x0_<, t_, x_D := 0.5 3 c0�a HExp@- Sqrt@c0D�2 Hx - x0 - c0 tL^2DL;s@8a_, c0_, x0_<, t_, x_D := 3 c0�a HSech@ Sqrt@c0D�2 Hx�2 - x0�2 - c0 tLDL^2;phi = s@81, 3, 25<, 0, xD + 0 s@81, 3�2, 35<, 0, xD +

s@81, 3, 25 + cell<, 0, xD + 0 s@81, 3�2, 35 + cell<, 0, xD +

s@81, 3, 25 - cell<, 0, xD + 0 s@81, 3�2, 35 - cell<, 0, xD;Hphi �. x ® 0.L � Hphi �. x ® cellL

ss = NDSolve@kdv@81<, t, xD && p@0, xD � phi && p@t, 0D � p@t, cellD, p,8t, 0, 40<, 8x, 0, cell<, Method -> StiffnessSwitchingD; �� Timing

<< Graphics‘Animation‘

Table@Plot@Evaluate@p@i�4, xD �. First@ssDD, 8x, 0, cell<,PlotRange ® 8-2, 15<, ImageSize ® 600D, 8i, 0, 20<D;

112

Appendix B: Sketch of Lagrangian Mechanics

B.1 Variational principle

The principle of variational mechanics is that u(t) minimises an action S (hereif it were a particle in motion it would be more conventional to call the positionx(t)but this would be confusing here, since later we will define a u as a fieldtheory over x, t);

S[u] =

∫ t2

t1

L(u, ut) dt (441)

In general L can also depend explicitly on t but we won’t consider this here.L is called the Lagrangian , and is given by

L = T − V (442)

S is called the action . The boundary conditions are that (u(t1) and u(t2) arefixed.) For a particle in falling under gravity

L =m

2u2t −mgu (443)

To find the solution, if the functional S[u] minimizes the action, it shouldbe a variational minimum; let u→ u+ δu then if S → S + δS we have

δS = 0 =

∫L(u+ δu, ut + δut)− L(u, ut) dt

=

∫L(u, ut) +

∂L

∂uδu+

∂L

∂utδut − L(u, ut) dt

=

∫ (∂L

∂u− d

dt

∂L

∂ut

)δu dt (444)

where to get the last line I integrated by parts. This should be true for anyδu so that we get the Euler-Lagrange eqn’s

∂L

∂u− d

dt

∂L

∂ut= 0 ∀t (445)

For the example ∂L∂ut

= mut and ∂L∂u

= −mg so utt = −g.More generally for a non-relativistic particle moving in a potential V (u)

we get

L =m

2u2t − V (u) (446)

and hencemutt = −V ′(u) (447)

(The rate of gain of K.E. = rate of loss of P.E.) The variational principlereplaces equation of motion (Newton’s law in this case).

Note it’s T − V . The best “path” u(t) tries to share the energy equallybetween T and V as it goes

113

B.2 Variational principle in field theory

For our equations we have e.g. θ(x, t) and u(x, t) so that we have a quantity(field) defined over x, t. Assume x ∈ R and u(±∞) = constant. Certainlysufficient to get solitons plus much more. Now define the Lagrange density ,L such that

S[u] =

∫ t2

t1

∫ ∞

∞L(u, ut, ux) dxdt (448)

let u→ u+ δu then if S → S + δS we have

δS = 0 =

∫ ∫L(u+ δu, ut + δut, ux + δux)− L(u, ut, ux) dtdx

=

∫ ∫∂L∂u

δu+∂L∂ut

δut +∂L∂ux

δux dtdx

=

∫ ∫ (∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

)δu dtdx. (449)

Here we can get the last line by integrating by parts in the relevant integrand,x or t. (Note we are neglecting surface terms here which are important forconserved currents. In more general cases we would use Gauss’s theorem in 2dimensions to do the final step.) The E-L equations become

∂L∂u

− d

dt

∂L∂ut

− d

dx

∂L∂ux

= 0 (450)

B.3 Example - the Sine-Gordon equation

To get the SG equation from the variational principle define

T =ml2

2θ2t

V = mg(1− cos θ) +ml2

2θ2x (451)

Recall that θxx came from the linear force law of stretched elastic. Here θ2x

is therefore the contribution to the potential energy per unit length stored inthe elastic, above the “vacuum” energy. (Go back to the swinging pendulumpicture and convince yourself this is the case.)

Now take

L = T − V

=ml2

2θ2t −

ml2

2θ2x −mg(1− cos θ) (452)

The E-L equations δS = 0 then require subsituting

∂L∂θt

= ml2θt

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∂L∂θx

= −ml2θx

∂L∂θ

= −mgl sin θ

givingml2θtt −ml2θxx +mgl sin θ = 0 (453)

B.4 One dimensional relativistic field theory

We can define a relativistic field theory with a general potential P(u) by anal-ogy with the above. Define

T =1

2u2t

V =1

2u2x + P(u) (454)

and henceL = u2

t − u2x − P(u) (455)

The Euler-Lagrange equation 450 now give

utt − uxx + P ′ = 0 (456)

Note that a travelling solution of the form u = f(x− vt) yields the equation

f ′′ = γ2P ′ (457)

Integrating once gives1

2(f ′)2 − γ2P = A (458)

where A is a constant of integration. Again we see that any travelling wavesolution will be equivalent to a particle rolling in the inverse potential. Fromhere we could integrate again to find an implicit solution for f .

Note that we can also define the total energy density as we did for Bogo-molny;

E = T + V (459)

• In some cases (e.g. QM) the lagrangian is more fundamental (there isno equation of “motion” of quantum particles)

• It’s often quicker and more intuitive - e.g. Bogomolny

• Important later for conservation laws

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Appendix C: Lightcone coordinates

Also called characteristic or Null coordinates. In many relativistic cases it isuseful to define coordinates where light rays travel along the axes. Recall thatthe worldlines of light rays are on the lightcone given by x = ±t (where I’llset the speed of light = 1), as in the figure below;

x− x

x

t

+

Coordinates of any point (x, t) relative to the x± axes are given by a rotationthrough 45o;

x± =1

2(x± t) (460)

(By convention there is a factor√

2 with respect to a pure rotation which is

done by the matrix 1√2

(1 1−1 1

)).

Then differentiation can be done by

∂

∂x+

=∂t

∂x+

∂

∂t+

∂x

∂x+

∂

∂x=

∂

∂t+

∂

∂x∂

∂x−=

∂t

∂x−

∂

∂t+

∂x

∂x−

∂

∂x= − ∂

∂t+

∂

∂x(461)

where I used

t = x+ − x−

x = x+ + x− (462)

You can check that∂x±∂x±

= 1 ;∂x±∂x∓

= 0,

and also that

∂

∂x+∂x−=

(∂

∂x+∂

∂t

) (∂

∂x− ∂

∂t

)=

∂2

∂x2− ∂2

∂t2(463)

so thatuxx − utt = u+− (464)

where I’ll use subscripts +− for the differentiation by x+ and x− respectively.

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Appendix D: Commutators

Commutators are the Quantum Mechanical equivalent of Poisson brackets.They act on operators and are defined as follows

[A,B] = AB −BA (465)

where both A and B are operators, e.g. matrices, or differential operators suchas ∂x. If [A,B] = 0 then the two operators are said to commute, and theirordering is irrelevant. If two operators do not commute then the ordering isimportant. In other words, if [A,B] = C, then by definition

AB = BA+ C (466)

Using the above definition the following rules are useful in expanding andevaluating complicated expressions;

[A,A] = 0 ∀A[A,B + C] = [A,B] + [A,C]

[A+B,C +D] = [A,C] + [A,D] + [B,C] + [B,D]

[A,BC] = [A,B]C +B[A,C] (467)

These rules are not hard to derive. For example the last one follows by writingout explicitly

[A,BC] = ABC −BCA

= ABC −BAC +BAC −BCA

= (AB −BA)C +B(AC − CA)

= [A,B]C +B[A,C].

Be careful to maintain the ordering. Calling [A,B] = φ then the above rulesclearly imply

[An, B] = φAn−1 + AφAn−1 + . . .+ An−1φ

If also [φ,A] = 0 then it follows that

[f(A), B] = φf ′(A)

where f(A) is any differentiable function of A.From these rules more specific ones follow. For example if [A,B] = 0 then

[A, f(B)C] = f(B) [A,C]

where f(B)is any function of B. A final identity (which is useful in QMalthough probably not for solitons) is the famous Baker-Campbell-Hausdorfftheorem;

eAeB = eBeAe[A,B].

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This is easy to show to second order in A,B;

(1 + A+A2

2+ . . .)(1 +B +

B2

2+ . . .) = (1 +B +

B2

2. . .)(1 + A+

A2

2. . .)(1 + [A,B] + . . .)

1 + A+B + AB +A2

2+B2

2= 1 +B + A+

A2

2+B2

2+BA+ [A,B]

118